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ELEMENTS 



GEO-IETRY 

% 

AND 

CONIC SECTIONS. 

% 




BY ELIAS 



OOMIS, A.M., 


V| 

PROFESSOR OF MATHEMATICS AND NATURAL PHILOSOPHY IN THE UNIVERSITY OF THE CITY 
OF NEW YORK, AND AUTHOR OF A “ TREATISE ON ALGEBRA.” 


» 



NEW YORK: 

* 

HARPER & BROTHERS, PUBLISPIERS, 

82 CLIFF STREET. 

184 7 . 





Entered, according to Act of Congress, in the year one thousand 
eight hundred and forty-seven, by 

Elias Loomis, 

in the Clerk’s Office of the District Court of the Southern District 

of New York. 


TO THE 


HON. THEODORE FRELINGHUYSEN, LL.D, 

CHANCELLOR OF THE UNIVERSITY OF THE CITY OF NEW YORK, 


THE FRIEND OF EDUCATION, THE PATRIOT STATESMAN, 
AND THE CHRISTIAN PHILANTHROPIST, 


movk 


IS RESPECTFULLY DEDICATED 

BY 


THE AUTHOR 



PREFACE. 


In the following treatise, an attempt has been made to combine the 
peculiar excellencies of Euclid and Legendre. The Elements of Euclid 
have long been celebrated as furnishing the most finished specimens of 
logic ; and on this account they still retain their place in many seminaries 
of education, notwithstanding the advances which science has made in 
modern times. But the deficiencies of Euclid, particularly in Solid Ge¬ 
ometry, are now so palpable, that few institutions are content with a 
simple translation from the original Greek. The edition of Euclid 
chiefly used in this country, is that of Professor Playfair, who has sought, 
by additions and supplements, to accommodate the Elements of Euclid 
to the present state of the mathematical sciences. But, even with these 
additions, the work is incomplete on Solids, and is very deficient on 
Spherical Geometry. Moreover, the additions are often incongruous 
with the original text; so that most of those who adhere to the use of 
Playfair’s Euclid, will admit that something is still wanting to a perfect 
treatise. At most of our colleges, the work of Euclid has been super¬ 
seded by that of Legendre. It seems superfluous to undertake a defense 
of Legendre’s Geometry, when its merits are so generally appreciated. 
No one can doubt that, in respect of comprehensiveness and scientific 
arrangement, it is a great improvement upon the Elements of Euclid. 
Nevertheless, it should ever be borne in mind that, with most students 
in our colleges, the ultimate object is not to make profound mathemati¬ 
cians, but to make good reasoners on ordinaiy subjects. In order to 
secure this advantage, the learner should be trained, not merely to give 
the outline of a demonstration, but to state every part of the argument 
with minuteness and in its natural order. Now, although the model of 
Legendre is, for the most part, excellent, his demonstrations are 
often mere skeletons. They contain, indeed, the essential part of an 
argument; but the general student does not derive from them the high¬ 
est benefit which may accrue from the study of Geometry as an exercise 
in reasoning. 

While, then, in the following treatise, I have, for the most part, fol¬ 
lowed the arrangement of Legendre, I have aimed to give his demonstra¬ 
tions somewhat more of the logical method of Euclid. I have also made 



VI 


PREFACE. 


some changes in arrangement. Several of Legendre’s propositions have 
been degraded to the rank of corollaries, while some of his corollaries 
and scholiums have been elevated to the dignity of primary propositions. 
His lemmas have been proscribed entirely, and most of his scholiums 
have received the more appropriate title of corollary. The quadrature 
of the circle is developed in an order somewhat different from any thing 
I have elsewhere seen. The propositions are all enunciated in general 
terms, with the utmost brevity which is consistent with clearness ; and, 
in order to remind the student to conclude his recitation with the enun¬ 
ciation of the proposition, the leading words are repeated at the close of 
each demonstration. As the time given to mathematics in our colleges 
is limited, and a variety of subjects demand attention, no attempt has 
been made to render this a complete record of all the known propositions 
of Geometry. On the contrary, nearly every thing has been excluded 
which is not essential to the student’s progress through the subsequent 
parts of his mathematical course. 

Considerable attention has been given to the construction of the dia¬ 
grams. I have aimed to reduce them all to nearly uniform dimensions, 
and to make them tolerable approximations to the objects they were de¬ 
signed to represent. I have made free use of dotted lines. Generally, 
the black lines are used to represent those parts of a figure which are 
directly involved in the statement of the proposition ; while the dotted 
lines exhibit the parts which are added for the purposes of demonstration. 
In Solid Geometry the dotted lines commonly denote the parts which 
would be concealed by an opaque solid; while in a few cases, for pecul¬ 
iar reasons, both of these rules have been departed from. Throughout 
Solid Geometry the figures have generally been shaded, which addition, 
it is hoped, will obviate some of the difficulties of which students frequent¬ 
ly complain. 

As the study of Analytical Geometry has lately become somewhat 
popular, a geometrical treatise on Conic Sections may be thought to be 
behind the spirit of the age. The few pages here devoted to this sub¬ 
ject can, of course, be read or omitted at the discretion of the teacher. 
It is believed, however, that those who intend to pursue Analytical Ge¬ 
ometry, will be benefited by some previous acquaintance with the simpler 
properties of the Conic Sections; while those who have not time or in¬ 
clination for this study, will here find demonstrated those properties 
which are indispensable to the study of Astronomy. 




CONTENTS, 


PLANE GEOMETRY. 

BOOK i. 

General Principles. Pa g 

BOOK II. 

Ratio and Proportion.35 

BOOK III. 

The Circle, and the Measure of Angles.44 

BOOK IV. 

The Proportions of Figures ..57 

BOOK V. 

Problems relating to the preceding Books. 83 

BOOK VI. 

Regular Polygons, and the Area of the Circle. .98 


SOLID GEOMETRY. 

BOOK VII. 

Planes and Solid Angles.112 

BOOK VIII. 

Polyedrons.127 

BOOK IX. 

Spherical Geometry. 148 

BOOK X. 

The Three round Bodies. 166 


CONIC SECTIONS. 

Parabola. 177 

Ellipse. 188 

Hyperbola. 205 


N.B .—When reference is made to a Proposition in the same Book, only the 
number of the Proposition is given; but when the Proposition is found in a 
different Book, the number of the Book is also specified. 















ELEMENTS OF GEOMETRY. 


BOOK L 

GENERAL PRINCIPLES. 

Definitions. 

1. Geometry is that branch of Mathematics which treats 
of the properties of extension and figure. 

Extension has three dimensions, length, breadth, and thick¬ 
ness. 

2 . A line is that which has length, without breadth or 
thickness. 

The extremities of a line are called points. A point, there¬ 
fore, has position, but not magnitude. 

3 . A straight line is the shortest path from one point to 
another. 

4 . Every line which is neither a straight line, nor compo¬ 
sed of straight lines, is a curved line. 

E 

Thus, AB is a straight line, ACDB is a 
broken line, or one composed of straight ^ 
lines, and AEB is a curved line. 

C D 

5 . A surface is that which has length and breadth, without 
thickness. 

6 . A plane is a surface in which any two points being ta¬ 
ken, the straight line which joins them lies wholly in that sur¬ 
face. 

7 . Every surface which is neither a plane, nor composed 
of plane surfaces, is a curved surface. 

8 . A solid is that which has length, breadth, and thick¬ 
ness, and therefore combines the three dimensions of exten¬ 
sion. 

9 . When two straight lines meet together, their inclina- 






10 


GEOMETRY. 


tion, or opening, is called an angle. The point of meeting 
is called the vertex , and the lines are called the sides of the 
angle. 

If there is only one angle at a point, it may 
be denoted by a letter placed at the vertex, as 
the angle at A. 

But if several angles are at one point, any one of them is 
expressed by three letters, of which the middle one is the let¬ 
ter at the vertex. 

Thus, the angle which is contained by the 
straight lines BC, CD, is called the angle 
BCD, or DCB. 

c 

Angles, like other quantities, may be added, subtracted, 
multiplied, or divided. Thus, the angle BCD is the sum of 
the two angles BCE, ECD; and the angle ECD is the differ¬ 
ence between the two angles BCD, BCE. 

10 . When a straight line, meeting another straight line, 

makes the adjacent angles equal to one another, 
each of them is called a right angle , and the 
straight line which meets the other is called a 
perpendicular to it. _ 

11 . An acute angle is one which is less than a 
right angle. 





An obtuse angle is one which is greater 
than a right angle. 

12 . Parallel straight lines are such as are 
in the same plane, and which, being produced 
ever so far both ways, do not meet. 

13 . A. plane figure is a plane terminated on all sides by 
lines either straight or curved. 

If the lines are straight, the space they in¬ 
close is called a rectilineal figure, or polygon , 
and the lines themselves, taken together, form 
the perimeter of the polygon. 

14 . The polygon of three sides is the simplest of all, and is 
called a triangle; that of four sides is called a quadrilateral ; 
that of five, a pentagon ; that of six, a hexagon , &c. 













BOOK I. 


11 


15 . An equilateral triangle is one which has its 
three sides equal. 


An isosceles triangle is that which has only two 
sides equal. 


A scalene triangle is one which has three un 
equal sides. 


16 . A right-angled triangle is one which has 
a right angle. The side opposite the right an¬ 
gle is called the hypothenuse. 

An obtuse-angled triangle is one which has an obtuse an¬ 
gle. An acute-angled triangle is one which has three acute 
angles. 

17 . Of quadrilaterals, a square is that which has 
all its sides equal, and its angles right angles. 

A rectangle is that which has all its angles right 1“ 
angles, but has not all its sides equal. 

A rhombus is that which has all its sides 
equal, but its angles are not right angles. 

A parallelogram is that which has its op¬ 
posite sides parallel. 

A trapezoid is that which has only two sides 
parallel. 

c 

18 . The diagonal of a figure is a line 
which joins the vertices of two angles not 
adjacent to each other. 

Thus, AC, AD, AE are diagonals. 

19 . An equilateral polygon is one which has all its sides 
equal. An equiangular polygon is one which has all its an¬ 
gles equal. 

20 . Two polygons are mutually equilateral when they 
have all the sides of the one equal to the corresponding sides 
of the other, each to each, and arranged in the same order. 

Two polygons are mutually equiangular when they have 


























12 


GEOMETRY. 


all the angles of the one equal to the corresponding angles 
of the other, each to each, and arranged in the same order. 

In both cases, the equal sides, or the equal angles, are call¬ 
ed homologous sides or angles. 

21 . An axiom is a self-evident truth. 

22 . A theorem is a truth which becomes evident by a train 
of reasoning called a demonstration. 

A direct demonstration proceeds from the premises by a 
regular deduction. 

An indirect demonstration shows that any supposition con¬ 
trary to the truth advanced, necessarily leads to an absurd¬ 
ity. 

23. A problem is a question proposed which requires a so¬ 
lution. 

24. A postulate requires us to admit the possibility of an 
operation. 

25. A proposition is a general term for either a theorem, 
or a problem. 

One proposition is the converse of another, when the con¬ 
clusion of the first is made the supposition in the second. 

26. A corollary is an obvious consequence, resulting from 
one or more propositions. 

27. A scholium is a remark appended to a proposition. 

28. An hypothesis is a supposition made either in the enun¬ 
ciation of a proposition, or in the course of a demonstration. 


Axioms. 

1. Things which are equal to the same thing are equal to 
each other. 

2. If equals are added to equals, the wholes are equal. * 

3. If equals are taken from equals, the remainders are 
equal. 

4. If equals are added to unequals, the wholes are unequal. 

5. If equals are taken from unequals, the remainders are 
unequal. 

6 . Things which are doubles of the same thing are equal to 
each other. 

7. Things which are halves of the same thing are equal to 
each other. 

8. Magnitudes which coincide with each other, that is, 
which exactly fill the same space, are equal. 

9. The whole is greater than any of its parts. 

10. The whole is equal to the sum of all its parts. 

11. From one point to another only one straight line can 
be drawn. 


BOOK I. 13 

12 . Two straight lines, which intersect one another, can 
not both be parallel to the same straight line. 


Explanation of Signs. 


For the sake of brevity, it is convenient to employ, to some 
extent, the signs of Algebra in Geometry. Those chiefly em¬ 
ployed are the following: 

The sign = denotes that the quantities between which it 
stands are equal; thus, the expression A=B signifies that A 
is equal to B. 

The sign 4 * IS called plus, and indicates addition; thus, 
A-f-B represents the sum of the quantities A and B. 

The sign — is called minus, and indicates subtraction ; thus, 
A—B represents what remains after subtracting B from A. 

The sign X indicates multiplication ; thus, A X B denotes 
the product of A by B. Instead of the sign X, a point is 
sometimes employed ; thus, A.B is the same as AxB. The 
same product is also sometimes represented without any in¬ 
termediate sign, by AB; but this expression should not be 
employed when there is any danger of confounding it with 
the line AB. 

A parenthesis ( ) indicates that several quantities are to 
be subjected to the same operation; thus, the expression 
AX(B+C—D) represents the product of A by the quantity 
B+C—D. 


The expression 


A 

^ indicates the quotient arising from divi- 

B 


ding A by B. 

A number placed before a line or a quantity is to be re¬ 
garded as a multiplier of that line or quantity ; thus, 3AB de¬ 
notes that the line AB is taken three times; iA denotes the 
half of A. 

The square of the line AB is denoted by AB 2 ; its cube by 

AB 3 . 

The sign indicates a root to be extracted ; thus, ^2 de¬ 
notes the square root of 2 ; V A X B denotes the square root 
of the product of A and B. 


N.B .—The first six books treat only of plane figures, or fig- 
ures drawn on a plane surface. 



14 


GEOMETRY. 


PROPOSITION I. THEOREM. 

All right angles are equal to each other . 

Let the straight line CD 
be perpendicular to AB, and 
GH to EF; then, by defini¬ 
tion 10, each of the angles 
ACD, BCD, EGH, FGH, will 

be a right angle ; and it is to ^ C ^ E G ^ 
be proved that the angle ACD is equal to the angle EGH. 

Take the four straight lines AC, CB, EG, GF, all equal to 
each other; then will the line AB be equal to the line EF 
(Axiom 2). Let the line EF be applied to the line AB, 
so that the point E may be on A, and the point F on B; 
then will the lines EF, AB coincide throughout ; for other¬ 
wise two different straight lines might be drawn from one 
point to another, which is impossible (Axiom 11). More¬ 
over, since the line EG is equal to the line AC, the point G 
will fall on the point C; and the line EG, coinciding with 
AC, the line GH will coincide with CD. For, if it could 
have any other position, as CK, then, because the angle EGH 
is equal to FGH (Def. 10), the angle ACK must be equal to 
BCK, and therefore the angle ACD is less than BCK. But 
BCK is less than BCD (Axiom 9 ); much more, then, is ACD 
less than BCD, which is impossible, because the angle ACD 
is equal to the angle BCD (Def. 10); therefore, GH can not 
but coincide with CD, and the angle EGH coincides with 
the angle ACD, and is equal to it (Axiom 8). Therefore, all 
right angles are equal to each other. 




H 


PROPOSITION II. THEOREM. 

The angles which one straight line makes with another upon 
one side of it , are either two right angles , or are together equal 
to two right angles. 

Let the straight line AB make with CD, 
upon one side of it, the angles ABC, ABD; 
these are either two right angles, or are to¬ 
gether equal to two right angles. 

For if the angle ABC is equal to ABD, -- 

each of them is a right angle (Def. 10); but c B D 








BOOK I. 


15 



if not, suppose the line BE to be drawn from 
the point B, perpendicular to CD; then will 
each of the angles CBE, DBE be a right 
angle. Now the angle CBA is equal to the 
sum of the two angles CBE, EBA. To 
each of these equals add the angle ABD; C B D 
then the sum of the two angles CBA, ABD will be equal to 
the sum of the three angles CBE, EBA, ABD (Axiom 2). 
Again, the angle DBE is equal to the sum of the two angles 
DBA, ABE. Add to each of these equals the angle EBC; 
then will the sum of the two angles DBE, EBC be equal to 
the sum of the three angles DBA, ABE, EBC. Now things 
that are equal to the same thing are equal to each other 
(Axiom 1) ; therefore, the sum of the angles CBA, ABD is 
equal to the sum of the angles CBE, EBD. But CBE, EBD 
are two right angles; therefore ABC, ABD are together 
equal to two right angles. Therefore, the angles which one 
straight line, &c. 

Corollary 1 . If one of the angles ABC, ABD is a right 
angle, the other is also a right angle. 

Cor. 2. If the line DE is perpendicular to 
AB, conversely, AB will be perpendicular to 

DE. 

For, because DE is perpendicular to AB, 
the angle DCA must be equal to its adjacent 
angle DCB (Def. 10), and each of them must 
be a right angle. But since ACD is a right angle, its adja¬ 
cent angle, ACE, must also be a right angle (Cor. 1). Hence 
the angle ACE is equal to the angle ACD (Prop. I.), and AB 
is perpendicular to DE. 

Cor. 3. The sum of all the angles BAC, 

CAD, DAE, EAF, formed on the same 
side of the line BF, is equal to two right 
angles; for their sum is equal to that of 
the two adjacent angles BAD, DAF. 



proposition hi. theorem (Converse of Prop . II.). 

If at a point in a straight line , two other straight lines , upon 
the opposite sides of it , make the adjacent angles together equal 
to two right angles , these two straight lines are in one and the 
same straight line. 

At the point B, in the straight line AB, let the two straight 
lines BC, BD, upon the opposite sides of AB, make the adja¬ 
cent angles, ABC, ABD, together equal to two right angles; 






16 


GEOMETRY. 


» 

then will BD be in the same straight line 
with CB. 

For, if BD is not in the same straight 
line with CB, let BE be in the same 
straight line with it; then, because the _ 
straight line CBE is met by the straight c 
line AB, the angles ABC, ABE are together equal to two 
right angles (Prop. II.). But, by hypothesis, the angles ABC, 
ABD are together equal to two right angles; therefore, the 
sum of the angles ABC, ABE is equal to the sum of the an¬ 
gles ABC, ABD. Take away the common angle ABC, and 
the remaining angle ABE, is equal (Axiom 3 ) to the remain¬ 
ing angle ABD, the less to the greater, which is impossible. 
Hence BE is not in the same straight line with BC; and in 
like manner, it may be proved that no other can be in the same 
straight line with it but BD. Therefore, if at a point, &c. 



PROPOSITION IV. THEOREM. 

Two straight lines, which have two points common , coincide 
with each other throughout their whole extent , and form hut one 
and the same straight line. 

Let there be two straight lines, having 
the points A and B in common; these 
lines will coincide throughout their whole 
extent. 

It is plain that the two lines must co¬ 
incide between A and B, for otherwise 
there would be two straight lines between A and B, which 
is impossible (Axiom 11). Suppose, however, that, on being 
produced, these lines begin to diverge at the point C, one 
taking the direction CD, and the other CE. From the point 
C draw the line CF at right angles with AC; then, since 
ACD is a straight line, the angle FCD is a right angle (Prop. 
II., Cor. 1); and since ACE is a straight line, the angle FCE 
is also a right angle ; therefore (Prop. I.), the angle FCE 
is equal to the angle FCD, the less to the greater, which is 
absurd. Therefore, two straight lines which have, &c. 



PROPOSITION V. THEOREM. 

If two straight lines cut one another , the vertical or opposite 
angles are equal. 

Let the two straight lines, AB, CD, cut one another in the 





BOOK I. 


17 


point E ; then will the angle AEC be equal 
to the angle BED, and the angle AED to 
the angle CEB. 

For the angles AEC, AED, which the 
straight line AE makes with the straight 
line CD, are together equal to two right D 

angles (Prop. II.) ; and the angles AED, DEB, which the 
straight line DE makes with the straight line AB, are also 
together equal to two right angles ; therefore, the sum of the 
two angles AEC, AED is equal to the sum of the two angles 
AED, DEB. Take away the common angle AED, and the 
remaining angle, AEC, is equal to the remaining angle DEB 
(Axiom* 3 ). In the same manner, it may be proved that the 
angle AED is equal to the angle CEB. Therefore, if two 
straight lines, &c. 

Cor. 1 . Hence, if two straight lines cut one another, the 
four angles formed at the point of intersection, are together 
equal to four right angles. 

Cor. 2 . Hence, all the angles made by any number of 
straight lines meeting in one point, are together equal to four 
right angles. 


PROPOSITION VI. THEOREM. 

If two triangles have two sides , and the included angle of the 
one , equal to two sides and the included angle of the other , each 
to each , the two triangles will he equal , their third sides will he 
equal , and their other angles will be equal , each to each. 

Let ABC, DEF be two triangles, 
having the side AB equal to DE, 
and AC to DF, and also the angle 
A equal to the angle D; then will 
the triangle ABC be equal to the 
triangle DEF. 

For, if the triangle ABC is ap¬ 
plied to the triangle DEF, so that the point A may be on D, 
and the straight line AB upon DE, the point B will coincide 
with the point E, because AB is equal to DE; and AB, coin¬ 
ciding with DE, AC will coincide with DF, because the an¬ 
gle A is equal to the angle D. Hence, also, the point C will 
coincide with the point F, because AC is equal to DF. But 
the point B coincides with the point E; therefore the base 
BC will coincide with the base EF (Axiom II), and will be 
equal to it. Hence, also, the whole triangle ABC will coin¬ 
cide with the whole triangle DEF, and will be equal to it; 






18 


GEOMETRY. 


and the remaining angles of the one, will coincide with the 
remaining angles of the other, and be equal to them, viz.: the 
angle ABC to the angle DEF, and the angle ACB to the an¬ 
gle DFE. Therefore, if two triangles, &c. 


PROPOSITION VII. THEOREM. 

If two triangles hhve two angles , and the included side of the 
one f equal to two angles and the included side of the other , each 
to each , the two triangles will he equal , the other sides will he 
equal , each to each , and the third angle of the one to the third 
angle of the other . 

Let ABC, DEF be two A 

triangles having the angle j/\ 

B equal to E, the angle C 
equal to F, and the inclu- y' 

ded sides BC, EF equal to _ _ 

each other; then will the B CE F 

triangle ABC be equal to the triangle DEF. 

For, if the triangle ABC is applied to the triangle DEF, so 
that the point B may be on E, and the straight line BC upon 
EF, the point C will coincide with the point F, because BC 
is equal to EF. Also, since the angle B is equal to the an¬ 
gle E, the side BA will take the direction DE, and therefore 
the point A will be found somewhere in the line DE. And, 
because the angle C is equal to the angle F, the line CA will 
take the direction FD, and the point A will be found some¬ 
where in the line DF; therefore, the point A, being found at 
the same time in the two straight lines DE, DF, must fall at 
their intersection, D. Hence the two triangles ABC, DEF 
coincide throughout, and are equal to each other; also, the 
two sides AB, AC are equal to the two sides DE, DF, each 
to each, and the angle A to the angle D. Therefore, if two 
triangles, &c. 



PROPOSITION VIII. THEOREM. 

Any two sides of a triangle are together greater than the 
third side. 

Let ABC be a triangle; any two of its 
sides are together greater than the third 
side, viz.: the sides BA, AC are greater 
than the side BC ; AB, BC are greater than 
AC; and BC, CA are greater than AB. 






BOOK I. 


19 


For the straight line AB is the shortest path between the 
points A and B (Def. 3); hence AB is less than the sum of 
AC and BC. For the same reason, BC is less than the sum 
of AB and AC; and AC less than the sum of AB and BC. 
Therefore, any two sides, &c. 


PROPOSITION IX. THEOREM. 

If from a point within a triangle , two straight lines are 
drawn to the extremities of either side , their sum will be less 
than the sum of the other two sides of the triangle. 



Let the two straight lines BD, CD be 
drawn from D, a point within the triangle 
ABC, to the extremities of the side BC; 
then will the sum of BD and DC be less 
than the sum of BA, AC, the other two 
sides of the triangle. 

Produce BD until it meets the side AC B 
in E ; and, because two sides of a triangle are greater than 
the third side (Prop. VIII.), the two sides BA, AE of the tri¬ 
angle BAE are greater than BE. To each of these add 
EC; then will the sum of BA, AC be greater than the sum 
of BE, EC. Again, because the two sides CE, ED of the 
triangle CED are greater than CD, if DB be added to each, 
the sum of CE, EB will be greater than the sum of CD, DB. 
But it has been proved that the sum of BA, AC is greater 
than BE, EC ; much more, then, are BA, AC greater than BD, 
DC. Therefore, if from a point, &c. 


PROPOSITION X. THEOREM. 

The angles at the base of an isosceles triangle 
one another. 

Let ABC be an isosceles triangle, of which 
the side AB is equal to AC; then will the angle 
B be equal to the angle C. 

For, conceive the angle BAC to be bisected 
by the straight line AD ; then, in the two trian¬ 
gles ABD, ACD, two sides AB, AD, and the in¬ 
cluded angle in the one, are equal to the two 
sides AC, AD, and the included angle in the other; there¬ 
fore (Prop. VI.), the angle B is equal to the angle C. There¬ 
fore, the angles at the base, &c. 


are equal to 



BBC 





20 


GEOMETRY. 


Cor. 1 . Hence, also, the line BD is equal to DC, and the 
angle ADB equal to ADC; consequently, each of these an¬ 
gles is a right angle (Def. 10). Therefore, the line bisecting 
the vertical angle of an isosceles triangle bisects the base at 
right angles ; and, conversely, the line bisecting the base of an 
isosceles triangle at right angles bisects also the vertical angle. 

Cor. 2. Every equilateral triangle is also equiangular. 

Scholium. Any side of a triangle may be considered as 
its base f and the opposite angle as its vertex ; but in an isos¬ 
celes triangle, that side is usually regarded as the base,which 
is not equal to either of the others. 


proposition xi. theorem ( Converse of Prop. X.). 

If two angles of a triangle are equal to one another , the op¬ 
posite sides are equal , and the triangle is isosceles. 

Let ABC be a triangle having the angle 
ABC equal to the angle ACB; then will the 
side AB be equal to the side AC. 

For if AB is not equal to AC, one of them 
must be greater than the other. Let AB be 
the greater, and from it cut off DB equal to AC 
the less, and join CD. Then, because in the tri¬ 
angles DBC, ACB, DB is equal to AC, and BC 
is common to both triangles, also, by supposition, the angle 
DBC is equal to the angle ACB; therefore, the triangle DBC 
is equal to the triangle ACB (Prop. VI.), the less to the great¬ 
er, which is absurd. Hence AB is not unequal to AC, that 
is, it is equal to it. Therefore, if two angles, &c. 

Cor. Hence, every equiangular triangle is also equilateral. 



PROPOSITION XII. THEOREM. 

The greater side of every triangle is opposite to the greater 
angle; and , conversely , the greater angle is opposite to the 
greater side. 

Let ABC be a triangle, having the angle ACB 
greater than the angle ABC; then will the side 
AB be greater than the side AC. 

Draw the straight line CD, making the angle 
BCD equal to B; then, in the triangle CDB, the 
side CD must be equal to DB (Prop. XL). Add 
AD to each, then will the sum of AD and DC 



B 




BOOK I. 


21 


be equal to the sum of AD and DB. But AC is less than the 
sum of AD and DC (Prop. VIII.); it is, therefore, less than 
AB. 

Conversely , if the side AB is greater than the side AC, then 
will the angle ACB be greater than the angle ABC. 

For if ACB is not greater than ABC, it must be either 
equal to it, or less. It is not equal, because then the side 
AB would be equal to the side AC (Prop. XI.), which is con¬ 
trary to the supposition. Neither is it less, because then the 
side AB would be less than the side AC, according to the for¬ 
mer part of this proposition; hence ACB must be greater 
than ABC. Therefore, the greater side, &c. 


PROPOSITION XIII. THEOREM. 

If two triangles have two sides of the one equal to two sides 
of the other , each to each, but the included angles unequal, the 
base of that which has the greater angle,will be greater than 
the base of the other. 



Let ABC, DEF be two trian¬ 
gles, having two sides of the one 
equal to two sides of the other, 
viz.: AB equal to DE, and AC to 
DF, but the angle BAC greater 
than the angle EDF; then will 
the base BC be greater than the 
base EF. 

Of the two sides DE, DF, let DE be the side which is not 
greater than the other; and at the point D, in the straight 
line DE, make the angle EDG equal to BAC; make DG 
equal to AC or DF, and join EG, GF. 

Because, in the triangles ABC, DEG, AB is equal to DE, 
and AC to DG; also, the angle BAC is equal to the angle 
EDG; therefore, the base BC is equal to the base EG (Prop. 
VI.). Also, because DG is equal to DF, the angle DFG is 
equal to the angle DGF (Prop. X.). But the angle DGF is 
greater than the angle EGF; therefore the angle DFG is 
greater than EGF ; and much more is the angle EFG greater 
than the angle EGF. Now, in the triangle EFG, because 
the angle EFG is greater than EGF, and because the great¬ 
er side is opposite the greater angle (Prop. XII.), the side 
EG is greater than the side EF. But EG has been proved 
equal to BC ; and hence BC is greater than EF. Therefore, 
if two triangles, &c. 



22 


GEOMETRY. 


proposition xiv. theorem ( Converse of Prop . XIII.)• 

If two triangles have two sides of the one equal to two sides 
of the other , each to each , but the bases unequal , the angle con¬ 
tained by the sides of that which has the greater base , will be 
greater than the angle contained by the sides of the other . 

Let ABC, DEF be two triangles 
having two sides of the one equal to 
two sides of the other, viz.: AB equal 
to DE, and AC to DF, but the base 
BC greater than the base EF; then 
will the angle BAC be greater than 
the angle EDF. 

For if it is not greater, it must be 
either equal to it, or less. But the angle BAC is not equal 
to the angle EDF, because then the base BC would be equal 
to the base EF (Prop. VI.), which is contrary to the suppo¬ 
sition. Neither is it less, because then the base BC would be 
less than the base EF (Prop. XIII.), which is also contrary 
to the supposition; therefore, the angle BAC is not less than 
the angle EDF, and it has been proved that it is not equal 
to it; hence the angle BAC must be greater than the angle 
EDF. Therefore, if two triangles, &c. 



PROPOSITION XV. THEOREM. 

• 

If two triangles have the three sides of the one equal to the 
three sides of the other , each to each , the three angles will also 
be equal , each to each , and the triangles themselves will be 
equal. 


Let ABC, DEF be two trian¬ 
gles having the three sides of the 
one equal to the three sides of the 
other, viz.: AB equal to DE, BC 
to EF, and AC to DF; then will 
the three angles also be equal, 
viz.: the angle A to the angle D, 
the angle B to the angle E, and the angle C to the angle F. 

For if the angle A is not equal to the angle D, it must be 
either greater or less. It is not greater, because then the 
base BC would be greater than the base EF (Prop. XIII.), 
which is contrary to the hypothesis; neither is it less, be- 






BOOK I. 


23 


cause then the base BC would be less than the base EF 
(Prop. XIII.), which is also contrary to the hypothesis. 
Therefore, the angle A must be equal to the angle D. In 
the same manner, it may be proved that the angle B is equal 
to the angle E, and the angle C to the angle F ; hence the 
two triangles are equal. Therefore, if two triangles, &c. 

Scholium. In equal triangles, the equal angles are oppo¬ 
site to the equal sides; thus, the equal angles A and D are 
opposite to the equal sides BC, EF. 


PROPOSITION XVI. THEOREM. 

From a point without a straight line , only one perpendicular 
can he drawn to that line. 

Let A be the given point, and DE the 
given straight line; from the point A only 
one perpendicular can be drawn to DE. 

For, if possible, let there be drawn two _c 

perpendiculars AB, AC; produce the line 
AB to F, making BF equal to AB, and join 
CF. Then, in the triangles ABC, FBC, be¬ 
cause AB is equal to BF, BC is common to 
both triangles, and the angle ABC is equal to the angle FBC, 
being both right angles (Prop. II., Cor. 1); therefore, tw r o 
sides and the included angle of one triangle, are equal to two 
sides and the included angle of the other triangle; hence the 
angle ACB is equal to the angle FCB (Prop. VI.). But, 
since the angle ACB is, by supposition, a right angle, FCB 
must also be a right angle; and the two adjacent angles 
BCA, BCF, being together equal to two right angles, the two 
straight lines AC, CF must form one and the same straight 
line (Prop. III.); that is, between the two points A and F, 
two straight lines, ABF, ACF, may be drawn, which is im¬ 
possible (Axiom 11) ; hence AB and AC can not both be per¬ 
pendicular to DE. Therefore, from a point, &c. 

Cor. From the same point, C, in the 
line AB, more than one perpendicular to 
this line can not be drawn. For, if possi¬ 
ble, let CD and CE be two perpendicu¬ 
lars ; then, because CD is perpendicular 
to AB, the angle DCA is a right angle; a 
and, because CE is perpendicular to AB, 
the angle EC A is also a right angle. Hence, the angle ACD 
is equal to the angle ACE (Prop. I.), the less to the greater, 



A 



V 

w 






24 


GEOMETRY. 


which is absurd ; therefore, CD and CE can not both be per¬ 
pendicular to AB from the same point C. 


PROPOSITION XVII. THEOREM. 

If, from a point without a straight line , a perpendicular he 
drawn to this line , and oblique lines he drawn to different 
points: 

1st. The perpendicular will he shorter than any oblique line. 

2d. Two oblique lines , which meet the proposed line at equal 
distances from the perpendicular , will he equal. 

3 d. Of any two oblique lines , that which is further from the 
perpendicular will he the longer. 

Let DE be the given straight line, and 
A any point without it. Draw AB per¬ 
pendicular to DE; draw, also, the ob¬ 
lique lines AC, AD, AE. Produce the 
tine AB to F, making BF equal to AB, 
and join CF, DF. 

First. Because, in the triangles ABC, 

FBC, AB is equal to BF, BC is common 
to the two triangles, and the angle ABC is equal to the angle 
FBC, being both right angles (Prop. II., Cor. 1); therefore, 
two sides and the included angle of one triangle,are equal to 
two sides and the included angle of the other triangle ; hence 
the side CF is equal to the side CA (Prop. VI.). But the 
straight line ABF is shorter than the broken line ACF (Prop. 
VIII.); hence AB, tne half of ABF, is shorter than AC, the 
half of ACF. Therefore, the perpendicular AB is shorter 
than any oblique line, AC. 

Secondly. Let AC and AE be two oblique lines which 
meet the line DE at equal distances from the perpendicular; 
they will be equal to each other. For, in the triangles ABC, 
ABE, BC is equal to BE, AB is common to the two triangles, 
and the angle ABC is equal to the angle ABE, being both 
right angles (Prop. I.); therefore, two sides and the included 
angle of one triangle are equal to two sides and the included 
angle of the other ; hence the side AC is equal to the side 
AE (Prop. VI.). Wherefore, two oblique lines, equally dis¬ 
tant from the perpendicular, are equal. 

Thirdly. Let AC, AD be two oblique lines, of which AD 
is further from the perpendicular than AC; then will AD be 
longer than AC. For it has already been proved that AC is 
equal to CF; and in the same manner it may be proved that 
AD is equal to DF. Now, by Prop. IX., the sum of the two 





BOOK I. 


25 


lines AC, CF is less than the sum of the two lines AD, DF. 
Therefore, AC, the half of ACF, is less than AD, the half of 
ADF; hence the oblique line which is furthest from the per¬ 
pendicular is the longest. Therefore, if from a point, &c. 

Cor. 1 . The perpendicular measures the shortest distance 
of a point from a line, because it is shorter than any oblique 
line. 

Cor. 2. It is impossible to draw three equal straight lines 
from the same point to a given straight line. 


PROPOSITION XVIII. THEOREM. 


If through the middle point of a straight line a perpendic¬ 
ular is drawn to this line: 

1st. Every point in the perpendicular is equally distant from 
the two extremities of the line. 

2d. Every point out of the perpendicular is unequally dis¬ 
tant from those extremities. 



Let the straight line EF be drawn perpen¬ 
dicular to AB through its middle point, C. 

First. Every point of EF is equally dis¬ 
tant from the extremities of the line AB ; for, 
since AC is equal to CB, the two oblique 
lines AD, DB are equally distant from the 
perpendicular, and are, therefore, equal (Prop. 

XVII.). So, also, the two oblique lines AE, 

EB are equal, and the oblique lines AF, FB 
are equal; therefore, every point of the per¬ 
pendicular is equally distant from the extremities A and B. 

Secondly. Let I be any point out of the perpendicular. 
Draw the straight lines IA. IB; one of these lines must cut 
the perpendicular in some point, as D. Join DB; then, by 
the first case, AD is equal to DB. To each of these equals 
add ID, then will IA be equal to the sum of ID and DB. 
Now, in the triangle IDB, IB is less than the sum of ID and 
DB (Prop. VIII.); it is, therefore, less than IA ; hence, every 
point out of the perpendicular is unequally distant from the 
extremities A and B. Therefore, if through the middle 
point, &c. 

Cor. If a straight line have two points, each of which is 
equally distant from the extremities of a second line, it will 
be perpendicular to the second line at its middle point. 




26 


GEOMETRY. 


PROPOSITION XIX. THEOREM. 

If two right-angled triangles have the hypothenuse and a 
side of the one , equal to the hypothenuse and a side of the other , 
each to each , the triangles are equal 

Let ABC, DEF be two 
right-angled triangles, having 
the hypothenuse AC and the 
side AB of the one, equal to 
the hypothenuse DF and side 
DE of the other; then will 
the side BC be equal to EF, and the triangle ABC to the tri¬ 
angle DEF. • 

For if BC is not equal to EF, one of them must be greater 
than the other. Let BC be the greater, and from it cut off 
BG equal to EF the less, and join AG. Then, in the triangles 
ABG, DEF, because AB is equal to DE, BG is equal to EF, 
and the angle B equal to the angle E, both of them being 
right angles, the two triangles are equal (Prop. VI.), and AG 
is equal to DF. But, by hypothesis, AC is equal to DF, and 
therefore AG is equal to AC. Now the oblique line AC, be¬ 
ing further from the perpendicular than AG, is the longer 
(Prop. XVII.), and it has been proved to be equal, which is 
impossible. Hence BC is not unequal to EF, that is, it is equal 
to it; and the triangle ABC is equal to the triangle DEF 
(Prop. XV.). Therefore, if two right-angled triangles, &c. 



PROPOSITION XX. THEOREM. 


Two straight lines perpendicular to a third line , are par¬ 
allel. 


Let the two straight lines 
AC, BD be both perpendicu¬ 
lar to AB ; then is AC par¬ 
allel to BD. 

For if these lines are not 
parallel, being produced, they 


must meet on one side or the other of AB. Let them be pro¬ 
duced, and meet in O; then there will be two perpendicu¬ 
lars, OA, OB, let fall from the same point, on the same 
straight line, which is impossible (Prop. XVI.). Therefore, 
two straight lines, &c. • 









BOOK I. 


27 


PROPOSITION XXI. THEOREM. 

If a straight line, meeting two other straight lines , makes the 
interior angles on the same side, together equal to two right an¬ 
gles, the two lines are parallel. 

Let the straight line AB, which 
meets the two straight lines AC, BD, 
make the interior angles on the same 
side, BAC, ABD, together equal to two 
right angles; then is AC parallel to 
BD. 

From G, the middle point of the line 
AB, draw EGF perpendicular to AC ; it will also be perpen¬ 
dicular to BD. For the sum of the angles ABD and ABF is 
equal to two right angles (Prop. II.); and by hypothesis the 
sum of the angles ABD and BAC is equal to two right an¬ 
gles. Therefore, the sum of ABD and ABF is equal to the 
sum of ABD and BAC. Take away the common angle 
ABD, and the remainder, ABF, is equal to BAC; that is, 
GBF is equal to GAE. 

Again, the angle BGF is equal to the angle AGE (Prop. 
V.) ; and, by construction, BG is equal to GA ; hence the tri¬ 
angles BGF, AGE have two angles and the included side of 
the one, equal to two angles and the included side of the oth¬ 
er ; they are, therefore, equal (Prop. VII.) ; and the angle 
BFG is equal to the angle AEG. But AEG is, by construc¬ 
tion, a right angle, whence BFG is also a right angle; that 
is, tjie two straight lines EC, FD are perpendicular to the 
same straight line, and are consequently parallel (Prop. 
XX.). Therefore, if a straight line, &c. 

Scholium. When a straight line 
intersects two parallel lines, the in¬ 
terior angles on the same side , are 
those which lie within the parallels, A 
and on the same side of the secant 
line, as AGH, GHC; also, BGH, r 
GHD. 

Alternate angles lie within the 
parallels, on different sides of the 
secant line, and are not adjacent to each other, as AGH, 
GHD ; also, BGH, GHC. 

Either angle without the parallels being called an exterior 
angle, the interior and opposite angle on the same side, lies 
within the parallels, on the same side of the secant line, but 








28 


GEOMETRY. 


not adjacent; thus, GHD is an interior angle opposite to the 
exterior angle EGB; so, also, with the angles CHG, AGE. 


PROPOSITION XXII. THEOREM. 

If a straight line , intersecting two other straight lines , makes 
the alternate angles equal to each other , or makes an exterior 
angle equal to the interior and opposite upon the same side of 
the secant line , these two lines are parallel. 

Let the straight line EF, which 
intersects the two straight lines AB, 

CD, make the alternate angles AGH, 

GHD equal to each other; then AB A 
is parallel to CD. For, to each of 
the equal angles AGH, GHD, add c _ 
the angle HGB; then the sum of 
AGH and HGB will be equal to the 
sum of GHD and HGB. But AGH 
and HGB are equal to two right angles (Prop. II.); there¬ 
fore, GHD and HGB are equal to two right angles; and 
hence AB is parallel to CD (Prop. XXI.). 

Again, if the exterior angle EGB is equal to the interior 
and opposite angle GHD, then is AB parallel to CD. For, 
the angle AGH is equal to the angle EGB (Prop. V.); and, 
by supposition, EGB is equal to GHD; therefore the angle 
AGH is equal to the angle GHD, and they are alternate an¬ 
gles ; hence, by the first part of the proposition, AB is par¬ 
allel to CD. Therefore, if a straight line, &c. 



PROPOSITION XXIII. THEOREM. 

(<Converse of Propositions XXL and XXII.) 

If a straight line intersect two parallel lines , it makes the 
alternate angles equal to each other ; also , any exterior angle 
equal to the interior and opposite on the same side; and the 
two interior angles on the same side together equal to two right 
angles. 

Let the straight line EF intersect 
the two parallel lines AB, CD; the 
alternate angles AGH, GHD are 
equal to each other; the exterior an¬ 
gle EGB is equal to the interior and 
opposite angle on the same side, 

GHD ; and the two interior angles on 
the same side, BGH, GHD, are to¬ 
gether equal to two right angles. 







BOOK I. 


29 


For if AGH is not equal to GHD, through G draw the 
line KL, making the angle KGH equal to GHD; then KL 
must be parallel to CD (Prop. XXII.). But, by supposition, 
AB is parallel to CD; therefore, through the same point, G, 
two straight lines have been drawn parallel to CD, which is 
impossible (Axiom 12). Therefore, the angles AGH, GHD 
are not unequal, that is, they are equal to each other. Now 
the angle AGH is equal to EGB (Prop. V.), and AGH has 
been proved equal to GHD; therefore, EGB is also equal to 
GHD. Add to each of these equals the angle BGH; then 
will the sum of EGB, BGH be equal to the sum of BGH, 
GHD. But EGB, BGH are equal to two right angles (Prop. 
II.); therefore, also, BGH, GHD are equal to two right an¬ 
gles. Therefore, if a straight line, &c 

Cor. 1 . If a straight line is perpendicular to one of two 
parallel lines, it is also perpendicular to the other. 

Cor. 2. If two lines, KL and CD, make with EF the two 
angles KGH, GHC together less than two right angles, then 
will KL and CD meet, if sufficiently produced. 

For if they do not meet, they are parallel (Def. 12). But 
they are not parallel; for then the angles KGH, GHC would 
be equal to two right angles. 


PROPOSITION XXIV. THEOREM. 


Straight lines which are parallel to the same line , are paral¬ 
lel to each other . 

Let the straight lines AB, CD be 

each of them parallel to the line EF; e_ 

then will AB be parallel to CD. 

For, draw any straight line, as C-- 

PQR, perpendicular to EF. Then, 

since AB is parallel to EF, PR, which A- 

is perpendicular to EF, will also be 
perpendicular to AB (Prop. XXIII., Cor. 1); and since CD 
is parallel to EF, PR will also be perpendicular to CD. 
Hence, AB and CD are both perpendicular to the same 
straight line, and are consequently parallel (Prop. XX.). 
Therefore, straight lines which are parallel, &c. 


PROPOSITION XXV. THEOREM. 

Two parallel straight lines are every where equally distant 
from each other. 

Let AB, CD be two parallel straight lines. From any 






30 


GEOMETRY. 


points, E and F, in one of them, _ H G- 

draw the lines EG, FH perpendic- c 
ular to AB ; they will also be per¬ 
pendicular to CD (Prop. XXIII., — 

Cor. 1). Join EH; then, because A ^ E B 

EG and FH are perpendicular to the same straight line AB, 
they are parallel (Prop. XX.) ; therefore, the alternate an¬ 
gles, EHF, HEG, which they make with HE are equal 
(Prop. XXIII.). Again, because AB is parallel to CD, the 
alternate angles GHE, HEF are also equal. Therefore, the 
triangles HEF, EHG have two atigles of the one equal to 
two angles of the other, each to each, and the side EH inclu¬ 
ded between the equal angles, common; hence the triangles 
are equal (Prop. VII.); and the line EG, which measures the 
distance of the parallels at the point E, is equal to the line 
FH, which measures the distance of the same parallels at the 
point F. Therefore, two parallel straight lines, &c. 


PROPOSITION XXVI. THEOREM. 

Two angles are equal f when their sides are parallel , each to 
each , and are similarly situated. 

Let BAC, DEF be two angles, having 
the side BA parallel to DE, and AC to 
EF; the two angles are equal to each 
other. 

Produce DE, if necessary, until it meets 
AC in G. Then, because EF is parallel 
to GC, the angle DEF is equal to DGC 
(Prop. XXIII.); and because DG is par¬ 
allel to AB, the angle DGC is equal to BAC; hence the an¬ 
gle DEF is equal to the angle BAC (Axiom 1). Therefore, 
two angles, &c. 

Scholium. In this proposition, it is necessary that the two 
angles be similarly situated ; for, if we produce FE to H, the 
angle DEH has its sides parallel to those of the angle BAC; 
but the two angles are not equal. 



PROPOSITION XXVII. THEOREM. 

If one side of a triangle is produced , the exterior angle is 
equal to the sum of the two interior and opposite angles; and 
the three interior angles of every triangle are equal to two 
right angles. 

Let ABC be any plane triangle, and let the side BC be 










BOOK I. 


31 


produced to D ; then will the ex¬ 
terior angle ACD be equal to the 
sum of the two interior and oppo¬ 
site angles A and B ; and the sum 



A 


E 


of the three angles ABC, BCA, _V__ 

CAB is equal to two right angles. B CD 

For, conceive CE to be drawn parallel to the side AB of 
the triangle ; then, because AB is parallel to CE, and AC 
meets them, the alternate angles BAC, ACE are equal (Prop. 
XXIII.). Again, because AB is parallel to CE, and BD 
meets them, the exterior angle ECD is equal to the interior 
and opposite angle ABC. But the angle ACE was proved 
equal to BAC; therefore the whole exterior angle ACD is 
equal to the two interior and opposite angles CAB, ABC 
(Axiom 2). To each of these equals add the angle ACB; 
then will the sum of the two angles ACD, ACB be equal to 
the sum of the three angles ABC, BCA, CAB. ' But the an¬ 
gles ACD, ACB are equal to two right angles (Prop. II.); 
hence, also, the angles ABC, BCA, CAB are together equal 
to two right angles. Therefore, if one side of a triangle, &c. 

Cor. 1. If the sum of two angles of a triangle is given, the 
third may be found by subtracting this sum from two right 
angles. 

Cor. 2. If two angles of one triangle, are equal to two an¬ 
gles of another triangle, the third angles are equal, and the 
triangles are mutually equiangular. 

Cor. 3. A triangle can have but one right angle; for if 
there were two, the third angle would be nothing. Still less 
can a triangle have more than one obtuse angle. 

Cor. 4. In a right-angled triangle, the sum of the two acute 
angles is equal to one right angle. 

Cor. 5. In an equilateral triangle, each of the angles is one 
third of two right angles, or two thirds of one right angle. 


PROPOSITION XXVIII. THEOREM. 


The sum of all the interior angles of a polygon , is equal to 
twice as many right angles , wanting four y as the figure has 
sides. 

Let ABCDE be any polygon; then the sum of all its inte¬ 
rior angles A, B, C, D, E is equal to twice as many right an¬ 
gles, wanting four, as the figure has sides (see next page). 

For, from any point, F, within it, draw lines FA, FB, FC, 
&c., to all the angles. The polygon is thus divided into as 
many triangles as it has sides. Now the sum of the three 



32 


GEOMETRY. 



angles of each of these triangles, is equal 
to two right angles (Prop. XXVII.); 
therefore the sum of the angles of all the 
triangles, is equal to twice as many right 
angles as the polygon has sides. But 
the same angles are equal to the angles 
of the polygon, together with the angles 
at the point F, that is, together with four ■ A - 
right angles (Prop. V., Cor. 2). Therefore the angles of the 
polygon are equal to twice as many right angles as the fig¬ 
ure has sides, wanting four right angles. 

Cor. 1 . The sum of the angles of a quadrilateral is four 
right angles; of a pentagon, six right angles; of a hexagon, 
eight, &c. 

Cor. 2. All the exterior angles of a polygon are together 
equal to four right angles. Because every interior angle, ABC, 
together with its adjacent exterior an¬ 
gle, ABD, is equal to two right angles 
(Prop. II.) ; therefore the sum of all the 
interior and exterior angles, is equal to 
twice as many right angles as the poly¬ 
gon has sides ; that is, they are equal to 
all the interior angles of the polygon, 
together with four right angles. Hence jy 
the sum of the exterior angles must be 
equal to four right angles (Axiom 3). 



PROPOSITION XXIX. THEOREM. 


The opposite sides and angles of a parallelogram are equal 
to each other. 


Let ABDC be a parallelogram; then will A 
its opposite sides and angles be equal to v— 
each other. \ 

Draw the diagonal BC ; then, because AB \, 
is parallel to CD, and BC meets them, the C 


alternate angles ABC, BCD are equal to each other (Prop. 
XXIII.). Also, because AC is parallel to BD, and BC meets 
them, the alternate angles BCA, CBD are equal to each oth¬ 
er. Hence the two triangles ABC, BCD have two angles, 
ABC, BCA of the one, equal to two angles, BCD, CBD, of 
the other, each to each, and the side BC included between 
these equal angles, common to the two triangles; therefore 
their other sides are equal, each to each, and the third angle 






BOOK r. 


33 


the side AB to the side CD, and AC to BD, and the angle 
BAC equal to the angle BDC. Also, because the angle ABC 
is equal to the angle BCD, and the angle CBD to the angle 
BCA, the whole angle ABD is equal to the whole angle 
ACD. But the angle BAC has been proved equal to the an¬ 
gle BDC; therefore the opposite sides and angles of a par¬ 
allelogram are equal to each other. 

Cor. Two parallels, AB, CD, comprehended between two 
other parallels, AC, BD, are equal; and the diagonal BC di¬ 
vides the parallelogram into two equal triangles. 


proposition xxx. theorem ( Converse of Prop . XXIX.). 

If the opposite sides of a quadrilateral are equal, each to 
each, the equal sides are parallel, and the figure is a parallelo¬ 
gram. 


yf\ 


X 


Let ABDC be a quadrilateral, having its a 
opposite sides equal to each other, viz.: the V 
side AB equal to CD, and AC to BD; then 
will the equal sides be parallel, and the fig¬ 
ure will be a parallelogram. ^ " 

Draw the diagonal BC; then the triangles ABC, BCD 
have all the sides of the one equal to the corresponding sides 
of the other, each to each; therefore the angle ABC is equal 
to the angle BCD (Prop. XV.), and, consequently, the side 
AB is parallel to CD (Prop. XXII.). For a like reason, AC 
is parallel to BD; hence the quadrilateral ABDC is a par¬ 
allelogram. Therefore, if the opposite sides, &c. 


PROPOSITION XXXI. THEOREM. 

If two opposite sides of a quadrilateral are equal and par¬ 
allel, the other two sides are equal and parallel, and the figure 
is a parallelogram. 

Let ABDC be a quadrilateral, having the 
sides AB, CD equal and parallel; then will 
the sides AC, BD be also equal and parallel, 
and the figure will be a parallelogram. 

Draw the diagonal BC; then, because 
AB is parallel to CD, and BC meets them, the alternate an¬ 
gles ABC, BCD are equal (Prop. XXIII). Also, because 
AB is equal to CD, and BC is common to the two triangles 
ABC, BCD, the two triangles ABC, BCD have two sides and 







34 


GEOMETRY. 


the included angle of the one, equal to two sides and the in¬ 
cluded angle of the other; therefore, the side AC is equal 
to BD (Prop. VI.), and the angle ACB to the angle CBD. 
And, because the straight line BC meets* the two straight 
lines AC, BD, making the alternate angles BCA, CBD equal 
to each other, AC is parallel to BD (Prop. XXII.); hence 
the figure ABDC is a parallelogram. Therefore, if two op¬ 
posite sides, &c. 


PROPOSITION XXXII. THEOREM. 

The diagonals of every parallelogram bisect each other . 

Let ABDC be a parallelogram whose di- A 
agonals, AD, BC, intersect each other in E ; 
then will AE be equal to ED, and BE to 
EC. 

Because the alternate angles ABE, ECD 
are equal (Prop. XXIII.), and also the alternate angles EAB, 
EDC, the triangles ABE, DCE have two angles in the one 
equal to two angles in the other, each to each, and the inclu¬ 
ded sides AB, CD are also equal; hence the remaining sides 
are equal, viz.: AE to ED, and CE to EB. Therefore, the 
diagonals of every parallelogram, &c. 

Cor. If the side AB is equal to AC, the triangles AEB, 
AEC have all the sides of the one equal to the corresponding 
sides of the other, and are consequently equal; hence the 
angle AEB will equal the angle AEC, and therefore the di¬ 
agonals of a rhombus bisect each other at right angles . 





BOOK II. 


35 


BOOK II. 

RATIO AND PROPORTION. 

On the Relation of Magnitudes to Numbers . 

The ratios of magnitudes may be expressed by numbers 
either exactly or approximately; and in the latter case, the 
approximation can be carried to any required degree of pre¬ 
cision. 

Thus, let it be proposed to find the numerical ratio of two 
straight lines, AB and CD. 

From the greater line AB, cut A 
off a part equal to the less, CD, , , 

as many times as possible; for 
example, twice, with a remain- ? ? 

der EB. From CD, cut off a - 

part equal to the remainder EB as often as possible; for ex¬ 
ample, once, with a remainder FD. From the first remain¬ 
der, BE, cut off a part equal to FD as often as possible; for 
example, once, with a remainder GB. From the second re¬ 
mainder, FD, cut off a part equal to the third, GB, as many- 
times as possible. Continue this process until a remainder is 
found which is contained an exact number of times in the 
preceding one. This last remainder will be the common 
measure of the proposed lines; and regarding it as the meas¬ 
uring unit, we may easily find the values of the preceding 
remainders, and at length those of the proposed lines; whence 
we obtain their ratio in numbers. 

For example, if we find GB is contained exactly twice in 
FD, GB will be the common measure of the two proposed 
lines. Let GB be called unity, then FD will be equal to 2. 
But EB contains FD once,plus GB; therefore, EB=3. CD 
contains EB once,plus FD ; therefore, CD=5. AB contains 
CD twice, plus EB ; therefore, AB = 13. Consequently, the 
ratio of the two lines AB, CD is that of 13 to 5. 

However far the operation is continued, it is possible that 
we may never find a remainder which is contained an exact 
number of times in the preceding one. In such cases, the ex - 




36 


GEOMETRY. 


□ 


act ratio can not be expressed in numbers ; but, by taking the 
measuring unit sufficiently small, a ratio may always be 
found, which shall approach as near as we please to the true 
ratio. 

So, also, in comparing two sur- 
faces, we seek some unit of meas¬ 
ure which is contained an exact 
number of times in each of them. 

Let A and B represent two sur¬ 
faces, and let a square inch be 
the unit of measure. Now, if 
this measuring unit is contained 
15 times in A and 24 times in B, then the ratio of A to B is 
that of 15 to 24. And although it may be difficult to find 
this measuring unit, we may still conceive it to exist; or, if 
there is no unit which is contained an exact number of times 
in both surfaces, yet, since the unit may be made as small as 
we please, we may represent their ratio in numbers to any 
degree of accuracy required. 

Again, if we wish to find the ratio of two solids, A and B, 
we seek some unit of measure which is contained an exact 
number of times in each of them. If we take a cubic inch 
as the unit of measure, and we find it to be contained 9 times 
in A, and 13 times in B, then the ratio of A to B is the same 
as that of 9 to 13. And even if there is no unit which is 
contained an exact number of times in both solids, still, by 
taking the unit sufficiently small, we may represent their ra¬ 
tio in numbers to any required degree of precision. 

Hence the ratio of two magnitudes in geometry, is the 
same as the ratio of two numbers, and thus each magnitude 
has its numerical representative. We therefore conclude 
that ratio in geometry is essentially the same as in arith¬ 
metic, and we might refer to our treatise on algebra for such 
properties of ratios as we have occasion to employ. How¬ 
ever, in order to render the present treatise complete in it¬ 
self, we will here demonstrate the most useful properties. 


Definitions . 

Def. 1. Ratio is the relation which one magnitude bears to 
another with respect to quantity. 

Thus, the ratio of a line two inches in length, to another 
six inches in length is denoted by 2 divided by 6, i. e., f or 
i, the number 2 being the third part of 6. So, also, the ra¬ 
tio of 3 feet to 6 feet is expressed by f or ±. 

A ratio is most conveniently written as a fraction; thus, 
























BOOK II. 


37 


the ratio of A to B is written The two magnitudes com¬ 
pared together are called the terms of the ratio; the first is 
called the antecedent , and the second the consequent 

Def. 2. Proportion is an equality of ratios. 

Thus, if A has to B the same ratio that C has to D, these 
four quantities form a proportion, and we write it 
A_C 
B~D’ 

or A : B : : C : D. 

The first and last terms of a proportion are called the two 
extremes , and the second and third terms the two means. 

Of four proportional quantities, the last is called a fourth 
proportional to the other three, taken in order. 

Since ———» 

B D 

it is obvious that if A is greater than B, C must be greater 
than D ; if equal, equal; and if less, less ; that is, if one ante¬ 
cedent is greater than its consequent, the other antecedent 
must be greater than its consequent; if equal, equal; and if 
less, less. 

Def. 3. Three quantities are said to be proportional, when 
the ratio of the first to the second is equal to the ratio of the 
second to the third; thus, if A, B, and C are in proportion, 
then 

A : B : : B : C. 

In this case the middle term is said to be a mean propor¬ 
tional between the other two. 

Def. 4. Two magnitudes are said to be equimultiples of 
two others, when they contain those others the same number 
of times exactly. Thus, 7A, 7B are equimultiples of A and 
B ; so, also, are mA and mB. 

Def. 5. The ratio of B to A is said to be the reciprocal of 
the ratio of A to B. 

Def. 6. Inversion is when the antecedent is made the con¬ 
sequent, and the consequent the antecedent. 

Thus, if A : B : : C : D; 

then, inversely, 

B : A : : D : C. 

Def. 7. Alternation is when antecedent is compared with 
antecedent, and consequent with consequent. 

Thus, if A : B : : C : D; 

then, by alternation, 

A : C : : B : D. 

Def. 8. Composition is when the sum of antecedent and 
consequent is compared either with the antecedent or con¬ 
sequent. 


38 


GEOMETRY. 


Thus, if A : B :: C : D; 

then, by composition, 

A-fB : A : : C+D : C, and A+B : B : : C-fD : D. 

Def. 9. Division is when the difference of antecedent and 
consequent is compared either with the antecedent or con¬ 
sequent. 

Thus, if A : B : : C : D; 

then, by division, 

A—B : A : : C—D : C, and A—B : B : : C—D : D. 


Axioms. 

1. Equimultiples of the same, or equal magnitudes, are 
equal to each other. 

2. Those magnitudes of which the same or equal magni¬ 
tudes are equimultiples, are equal to each other. 


PROPOSITION I. THEOREM. 


If four quantities are proportional, the product of the two ex¬ 
tremes is equal to the product of the two means. 


It has been shown that the ratio of two magnitudes, wheth¬ 
er they are lines, surfaces, or solids, is the same as that of 
two numbers, which we call their numerical representatives. 

Let, then, A, B, C, D be the numerical representatives of 
four proportional quantities, so that A : B : : C : D ; then 
will AxD=BxC. 

For, since the four quantities are proportional, 

A C 
B = D 

Multiplying each of these equal quantities by B (Axiom 1), 
we obtain 


A = 


BxC 
D # 


Multiplying each of these last equals by D, we have 
AxD=BxC. 

Cor. If there are three proportional quantities , the product 
of the two extremes is equal to the square of the mean. 

Thus, if A : B : : B : C; 

then, by the proposition, 

A X C = B X B, which is equal to B\ 



BOOK II. 


39 


proposition ii. theorem (Converse of Prop. I.). 

If the product of two quantities is equal to the product of two 
other quantities , the first two may he made the extremes y and 
the other two the means of a proportion. 

Thus, suppose we have AxD=BxC; then will 
A : B : : C : D. 

For, since AxD=BxC, dividing each of these equals by 
D (Axiom 2), we have 


Dividing each of these last equals by B, we obtain 
A C 
B“D’ 

that is, the ratio of A to B is equal to that of C to D, 
or, A : B : : C : D. 


PROPOSITION III. THEOREM. 


If four quantities are proportional , they are also proportion¬ 
al when taken alternately. 

Let A, B, C, D be the numerical representatives of four 
proportional quantities, so that A : B : : C : D ; then will 
A : C : : B : D. 

For, since A : B : : C : D, 

by Prop. I., AxD=BxC. 

And, since A X D = B X C, 

by Prop. II., A : C :: B : D. 


PROPOSITION IV. THEOREM. 

Ratios that are equal to the same ratio t are equal to each 
other. 


Let 

A : B : 

: : C : D, 

and 

A : B : 

: : E : F; 

then will 

C : D : 

: : E : F. 

For, since 

A : B : 

:C:D, 


A C 
B“D* 


we have 



40 


GEOMETRY. 


And, since A : B : : E : F, 

. A E 

we have ~ 


But p and p being severally equal to g, must be equal to 

each other, and therefore 

C : D : : E: F. 

Cor. If the antecedents of one proportion are equal to the 
antecedents of another proportion, the consequents are pro¬ 
portional. 

If A : B : : C : D, 

and A : E : : C : F; 

then will B : D : : E : F. 

For, by alternation (Prop. III.), the first proportion be¬ 
comes 

A : C : : B : D, 
and the second, A : C : : E : F. 

Therefore, by the proposition, 

B : D : : E : F. 


PROPOSITION V. THEOREM. 


If four quantities are proportional , they are also proportion - 
al when taken inversely . 


Let 

then will 
For, since 
by Prop. I., 
or, 


therefore, by Prop. II., 


A : B : : C : D; 
B : A : : D : C. 
A : B : : C : D, 
AxD=BxC, 

B xC=AxD; 


B : A : : D : C. 


PROPOSITION VI. THEOREM. 


If four quantities are proportional , they are also proportion¬ 
al hy composition . 

Let A : B : : C : D, 

then will A+B : A : : C+D : C. 

For, since A : B : : C : D, 

by Prop. I., BxG=AxD. 

To each of these equals add 

AxC=AxC, 

then AxC-f BxC=AxC-f AxD, 


BOOK II. 


41 


or, (A+B) X C = A X (C+D). 

Therefore, by Prop. II., 

A+B : A : : C+D : C. 


PROPOSITION VII. THEOREM. 

If four quantities are proportional , they are also proportion¬ 
al by division . 

Let A : B : : C : D; 

then will A—B : A : : C—D : C. 

For, since A : B : : C : D, 

by Prop. I., BxC=AxD. 

Subtract each of these equals from AxC; 
then AxC—BxC=AxC—AxD, 

or, (A—B) x C=A x (C —D). 

Therefore, by Prop. II., 

A—B : A : : C—D : C. 

Cor . A+B : A—B : : C+D : C—D. 


PROPOSITION VIII. THEOREM. 

Equimultiples of two quantities have the same ratio as the 
quantities themselves . 

Let A and B be any two quantities, and mA, mB their 
equimultiples; then will 

A : B : : mA : mB. 

For mxAxB=mxAxB, 

or, AxmB=B x mA. 

Therefore, by Prop. II., 

A : B : : mA : mB. 


PROPOSITION IX. THEOREM. 


If any number of quantities are proportional , any one ante¬ 
cedent is to its consequent, as the sum of all the antecedents , is 
to the sum of all the consequents. 


Let A : B : 

then will A : B : 

For, since 
we have 
And, since 
we have 

To these equals add 


: : C : D : : E : F, &c.; 
: : A+C+E : B+D+F. 
A : B : : C : D, 
AxD=BxC. 

A : B : : E : F, 
AxF=BxE. 

AxB=AxB, 


42 


GEOMETRY. 


and we have 

AxB+AxD+AxF=AxB+BxC+BxE; 
or, A x (B+D+F)=B x (A+C+E). 

Therefore, by Prop. II., 

A : B : : A+C+E : B+D+F. 


PROPOSITION X. THEOREM. 


If four quantities are proportional , their squares or cubes 
are also proportional. 


Let 

then will 
and 

For, since 
by Prop. I., 
or, multiplying 
have 


A : B : : C : D; 

A a : B a : : C a : B a , 

A 3 : B 3 : : C 3 : D s . 

A : B : : C : D, 

AxD=BxC; 

each of these equals by itself (Axiom 1), we 


A a xD a =B a xC a ; 

and multiplying these last equals by AxD = BxC, we have 
A 3 xD 3 =B 3 xC\ 


Therefoie, by Prop. II., 

A a : B a : : C a : D a , 
and A 3 : B 3 : : C 3 : D 3 . 


PROPOSITION XI. THEOREM. 

If there are two sets of proportional quantities , the products 
of the corresponding terms are proportional. 

Let A : B : : C : D, 

and E : F : : G : H; 

then will AxE : BxF : : CxG : DxH. 

For, since A : B : : C : D, 

by Prop. I., AxD = BxC. 

And, since E : F : : G : H, 

by Prop. I., ExH=FxG. 

Multiplying together these equal quantities, we have 
AxDxExH=BxCxFxG; 
or, (AxE)x(DxH)=(BxF)x(CxG); 

therefore, by Prop. II., 

AxE: BxF :: CxG: DxH. 

Cor. If A : B: : C : D, 

and B : F: : G : H ; 

then A :F:: CxG: DxH. 



BOOK II. 


43 


For, by the proposition, 

AxB : BxF : : CxG : DxH. 

Also, by Prop. VIII., 

AxB : BxF : : A : F; 

hence, by Prop. IV., 

A : F::CxG:DxH. 


PROPOSITION XII. THEOREM. 


If three quantities are proportional , the first is to the third , 
as the square of the first to the square of the second . 


Thus, if A : B : : B :C; 

then A : C : : A 2 : B 2 . 

For, since A : B : : B : C, 

and A : B : : A : B ; 

therefore, by Prop. XI., 

A 2 : B a : ; A x B : B X C. 


But, by Prop. VIII., 

AxB:BxC::A:C; 
hence, by Prop. IV., A : C : : A 2 : B 2 . 


44 


GEOMETRY. 


BOOK III. 

THE CIRCLE, AND THE MEASURE OF ANGLES. 

Definitions. 

1. The circumference of a circle is a curved line, every 
point of which is equally distant from a point within, called 
the center. 

The circle is the space bounded by this 
curved line. 

2. A radius of a circle is a straight line 
drawn from the center to the circumference. 

A diameter of a circle is a straight line 
passing through the center, and terminated 
both ways by the circumference. 

Cor. All the radii of a circle are equal; all the diameters 
are equal also, and each double of the radius. 

3. An arc of a circle is any part of the circumference. 

The chord of an arc is the straight line which joins its two 

extremities. 

4. A segment of a circle is the figure included between an 
arc and its chord. 

5. A sector of a circle is the figure included between an 
arc, and the two radii drawn to the extremities of the arc. 

6. A straight line is said to be inscribed in a circle, when 
its extremities are in the circumference. 

An inscribed angle is one whose sides are 
inscribed. 

7. A polygon is said to be inscribed in a 
circle, when all its sides are inscribed. The 
circle is then said to be described about the 
polygon. 

8. A secant is a line which cuts the cir¬ 
cumference, and lies partly within and partly without the 
circle. 

9. A straight line is said to touch a circle, when it meets 
the circumference, and, being produced, does not cut it. 
Such a line is called a tangent , and the point in which it 
meets the circumference, is called the point of contact . 








BOOK III. 


45 


10. Two circumferences touch each 
other when they meet, but do not cut 
one another. 



11. A polygon is described about a circle, 
when each side of the polygon touches the cir¬ 
cumference of the circle. 

In the same case, the circle is said to be in¬ 
scribed in the polygon. 



PROPOSITION I. THEOREM. 

Every diameter divides the circle and its circumference into 
two equal parts. 

Let ACBD be a circle, and AB its di¬ 
ameter. The line AB divides the circle 
and its circumference into two equal parts. 

For, if the figure ADB be applied to the 
figure ACB, while the line AB remains 
common to both, the curve line ACB must 
coincide exactly with the curve line ADB. 

For, if any part of the curve ACB were to 
fall either within or without the curve ADB, there would be 
points in one or the other unequally distant from the center, 
which is contrary to the definition of a circle. Therefore, 
every diameter, &c. 



PROPOSITION IT. THEOREM. 


A straight line can not meet the circumference of a circle in 
more than two points. 

For, if it is possible, let the straight 
line ADB meet the circumference CDE 
in three points, C, D, E. Take F, the 
center of the circle, and join FC, FD, 

FE. Then, because F is the center of 
the circle, the three straight lines FC, 

FD, FE are all equal to each other; 
hence, three equal straight lines have 
been drawn from the same point to the same straight line, 





46 


GEOMETRY. 


which is impossible (Prop. XVII., Cor. 2, Book I.). There¬ 
fore, a straight line, &c. 


PROPOSITION III. THEOREM. 

In equal circles , equal arcs are subtended by equal chords; 
and , conversely , equal chords subtend equal arcs. 

Let ADB, EHF be 
equal circles, and let the 
arcs AID, EMH also be 
equal; then will the 
chord AD be equal to 
the chord EH. 

For, the diameter AB 
being equal to the diameter EF, the semicircle ADB may be 
applied exactly to the semicircle EHF, and the curve line 
AIDB will coincide entirely with the curve line EMHF 
(Prop. I.). But the arc AID is, by hypothesis, equal to the 
arc EMH; hence the point D will fall on the point H, and 
therefore the chord AD is equal to the qhord EH (Axiom 
11, B. I.). 

Conversely , if the chord AD is equal to the chord EH, then 
the arc AID will be equal to the arc EMH. 

For, if the radii CD, GH are drawn, the two triangles 
ACD, EGH will have their three sides equal, each to each, 
viz.: AC to EG, CD to GH, and AD equal to EH ; the tri¬ 
angles are consequently equal (Prop. XV., B. I.), and the an¬ 
gle ACD is equal to the angle EGH. Let, now, the semicir¬ 
cle ADB be applied to the semicircle EHF, so that AC may 
coincide with EG; then, since the angle ACD is equal to the 
angle EGH, the radius CD will coincide with the radius GH, 
and the point D with the point H. Therefore, the arc AID 
must coincide with the arc EMH, and be equal to it. Hence, 
in equal circles, &c. 



PROPOSITION IV. THEOREM. 

In equal circles , equal angles at the center , are subtended by 
equal arcs ; and , conversely , equal arcs subtend equal angles at 
the center. 

Let AGB, DHE be two equal circles, and let ACB, DFE 
be equal angles at their centers; then will the arc AB be 
equal to the arc DE. Join AB, DE ; and, because the cir- 




BOOK III. 


47 


cles AGB, DHE are equal, their 
radii are equal. Therefore, the 
two sides CA, CB are equal to 
the two sides FD, FE; also, the 
angle at C is equal to the angle 
at F; therefore, the base AB is 
equal to the base DE (Prop. VI., 

B. I.). And, because the chord AB 
is equal to the chord DE, the arc AB must be equal to the 
arc DE (Prop. III.). 

Conversely , if the arc AB is equal to the arc DE, the an¬ 
gle ACB will be equal to the angle DFE. For, if these an¬ 
gles are not equal, one of them is the greater. Let ACB be 
the greater, and take ACI equal to DFE; then, because 
equal angles at the center are subtended by equal arcs, the 
arc AI is equal to the arc DE. But the arc AB is equal to 
the arc DE; therefore, the arc AI is equal to the arc AB, 
the less to the greater, which is impossible. Hence the an¬ 
gle ACB is not unequal to the angle DFE, that is, it is equal 
to it. Therefore, in equal circles, &c. 



PROPOSITION y. THEOREM. 

In the same circle , or in equal circles , a greater arc is sub¬ 
tended by a greater chord; and , conversely , the greater chord 
subtends the greater arc. 

In the circle AEB, let the arc AE be 
greater than the arc AD; then will the 
chord AE be greater than the chord AD. 

Draw the radii CA, CD, CE. Now, if 
the arc AE were equal to the arc AD, 
the angle ACE would be equal to the an¬ 
gle ACD (Prop. IV.) ; hence it is clear 
that if the arc AE be greater than the arc 
AD, the angle ACE must be greater than the angle ACD. 
But the two sides AC, CE of the triangle ACE are equal to 
the two AC, CD of the triangle ACD, and the angle ACE is 
greater than the angle ACD; therefore, the third side AE is 
greater than the third side AD (Prop. XIII., B. I.); hence 
the chord which subtends the greater arc is the greater. 

Conversely , if the chord AE is greater than the chord AD, 
the arc AE is greater than the arc AD. For, because the 
two triangles ACE, ACD have two sides of the one equal 
to two sides of the other, each to each, but the base AE of 
the one is greater than the base AD of the other, therefore 






48 


GEOMETRY. 


the angle ACE is greater than the angle ACD (Prop. XIV., 
B. I.); and hence the arc AE is greater than the arc AD 
(Prop. IV.). Therefore, in the same circle, &c. 

Scholium. The arcs here treated of are supposed to be 
less than a semicircumference. If they were greater, the op¬ 
posite property would hold true, that is, the greater the arc 
the smaller the chord. 


PROPOSITION VI. THEOREM. 

The radius which is perpendicular to a chord, bisects the 
chord , and also the arc which it subtends. 

Let ABG be a circle, of which AB is a 
chord, and CE a radius perpendicular to 
it; the chord AB will be bisected in D, 
and the arc AEB will be bisected in E. 

Draw the radii CA, CB. The two right- 
angled triangles CDA, CDB have the side 
AC equal to CB, and CD common; there¬ 
fore the triangles are equal, and the base 
AD is equal to the base DB (Prop. XIX., 

B. I.). 

Secondly , since ACB is an isosceles triangle, and the line 
CD bisects the base at right angles, it bisects also the verti¬ 
cal angle ACB (Prop. X., Cor. 1, B. I.). And, sinde the an¬ 
gle ACE is equal to the angle BCE, the arc AE must be 
equal to the arc BE (Prop. IV.) ; hence the radius CE, per¬ 
pendicular to the chord AB, divides the arc subtended by 
this chord, into two equal parts in the point E. Therefore, 
the radius, &c. 

Scholium. The center C, the middle point D of the chord 
AB, and the middle point E of the arc subtended by this 
chord, are three points situated in a straight line perpendic¬ 
ular to the chord. Now two points are sufficient to deter¬ 
mine the position of a straight line; therefore any straight 
line which passes through two of these points, will necessari¬ 
ly pass through the third, and be perpendicular to the chord. 
Hence the perpendicular to the middle of a chord, passes through 
the center of the circle , and through the middle of the arc sub¬ 
tended by the chord. 





BOOK III. 


49 


PROPOSITION VII. THEOREM. 

Through three given points, not in the same straight line, 
one circumference may he made to pass, and but one . 

Let A, B, C be three points not in the same straight line; 
they all lie in the circumference of the same circle. Join 
AB, AC, and bisect these lines by the 
perpendiculars DF, EF; DF and EF 
produced will meet one another. For, 
join DE; then, because the angles ADF, 

AEF are together equal to two right an- 
gles, the angles FDE and FED are to- B 
gether less than two right angles; there¬ 
fore DF and EF will meet if produced 
(Prop. XXIII., Cor. 2, B. I.). Let them 
meet in F. Since this point lies in the perpendicular DF, it is 
equally distant from the two points A and B (Prop. XYIII., 
B. I.); and, since it lies in the perpendicular EF, it is equally 
distant from the two points A and C; therefore the three 
distances FA, FB, FC are all equal; hence the circumfe¬ 
rence described from the center F with the radius FA will 
pass through the three given points A, B, C. 

Secondly. No other circumference can pass through the 
same points. For, if there were a second, its center could 
not be out of the line DF, for then it would be unequally dis¬ 
tant from A and B (Prop. XYIII., B. I.); neither could it be 
out of the line FE, for the same reason ; therefore, it must be 
on both the lines DF, FE. But two straight lines can not 
cut each other in more than one point; hence only one cir¬ 
cumference can pass through three given points. Therefore, 
through three given points, &c. 

Cor. Two circumferences can not cut each other in more 
than two points, for, if they had three common points, they 
would have the same center, and would coincide with each 
other. 



PROPOSITION VIII. THEOREM. 

Equal chords are equally distant from the center; and of two 
unequal chords, the shortest is furthest from the center. 

Let the chords AB, DE, in the circle ABED, be equal to 
one another; they are equally distant from the center. Take 

D 



50 


GEOMETRY. 


C, the center of the circle, and from it 
draw CF, CG, perpendiculars to AB, 

DE. Join CA, CD ; then, because the 
radius CF is perpendicular to the chord 
AB, it bisects it (Prop. VI.). Hence 
AF is the half of AB ; and, for the same 
reason, DG is the half of DE. But AB 
is equal to DE ; therefore AF is equal 
to DG (Axiom 7, B. I.). Now, in the 
right-angled triangles ACF, DCG, the hypothenuse AC is 
equal to the hypothenuse DC, and the side AF is equal to 
the side DG; therefore the triangles are equal, and CF is 
equal to CG (Prop. XIX., B. I.); hence the two equal chords 
AB, DE are equally distant from the center. 

Secondly. Let the chord iVH be greater than the chord DE ; 
DE is further from the center than AH. For, because the 
chord AH is greater than the chord DE, the arc ABH is 
greater than the arc DE (Prop. V.). From the arc ABH 
cut off a part, AB, equal to DE ; draw the chord AB, and 
let fall CF perpendicular to this chord, and Cl perpendicular 
to AH. It is plain that CF is greater than CK, and CK 
than Cl (Prop. XVII., B. I.); much more, then, is CF great¬ 
er than CL But CF is equal to CG, because the chords AB, 
DE are equal; hence CG is greater than CL Therefore, 
equal chords, &c. 

Cor. Hence the diameter is the longest line that can be in¬ 
scribed in a circle. 



PROPOSITION IX. THEOREM. 

A straight line perpendicular to a diameter at its extremity , 
is a tangent to the circumference. 

Let ABG be a circle, the center of which is C, and the di¬ 
ameter AB; and let AD be drawn from A perpendicular to 
AB ; AD will be a tangent to the circum¬ 
ference. D 

In AD take any point E, and join 
CE ; then, since CE is an oblique line, 
it is longer than the perpendicular CA 
(Prop. XVII., B. I.). Now CA is equal 
to CK ; therefore CE is greater than B 
CK, and the point E must be without 
the circle. But E is any point whatev¬ 
er in the line AD; therefore AD has 
only the point A in common with the 






BOOK III. 


51 


circumference, hence it is a tangent (Def. 9). Therefore, 
a straight line, &c. 

Scholium. Through the same point A in the circumfer¬ 
ence, only one tangent can be drawn. For, if possible, let a 
second tangent, AF, be drawn; then, since CA can not be 
perpendicular to AF (Prop. XVI., Cor., B. I.), another line, 
CH, must be perpendicular to AF, and therefore CH must be 
less than CA (Prop. XVII., B. I.; hence the point H falls 
within the circle, and AH produced will cut the circumfer¬ 
ence. 


PROPOSITION X. THEOREM. 

Two parallels intercept equal arcs on the circumference. 

The proposition admits of three cases: 

First. When the two parallels are se¬ 
cants, as AB, DE. Draw the radius CH 
perpendicular to A*B ; it will also be per¬ 
pendicular to DE (Prop. XXIII., Cor. 

1, B. I.) ; therefore, the point H will be 
at the same time the middle of the arc 
AHB, and of the arc DHE (Prop. VI.). 

Hence the arc DH is equal to the arc 
HE, and the arc AH equal to HB, and therefore the arc AD 
is equal to the arc BE (Axiom 3, B. I.). 

Second. When one of the two par¬ 
allels is a secant, and the other a tan- D 
gent. To the point of contact, H, . 
draw the radius CH; it will be per¬ 
pendicular to the tangent DE (Prop. 

IX.), and also to its parallel AB. But 
since CH is perpendicular to the chord 
AB, the point H is the middle of the 
arc AHB (Prop. VI.) ; therefore the F 
arcs AH, HB, included between the 
parallels AB, DE, are equal. 

Third. If the two parallels DE, FG are tangents, the one 
at H, the other at K, draw the parallel secant AB; then, ac¬ 
cording to the former case, the arc AH is equal to HB, and 
the arc AK is equal to KB; hence the whole arc HAK is 
equal to the whole arc HBK (Axiom 2, B. I.). It is also ev¬ 
ident that each of these arcs is a semicircumference. There¬ 
fore, two parallels, &c. 










52 


GEOMETRY. 


PROPOSITION XI. THEOREM. 


If two circumferences cut each other , the chord which joins 
the points of intersection , is bisected at right angles by the 
straight line joining their centers. 


Let two circum¬ 
ferences cut each 
other in the points A 
and B ; then will the 
line AB be a com¬ 
mon chord to the 
two circles. Now, if 
a perpendicular be 
erected from the middle of this chord, it will pass through C 
and D, the centers of the two circles (Prop. VI., Schol.). 
But only one straight line can be drawn through two given 
points; therefore, the straight line which passes through the 
centers, will bisect the common chord at right angles. 



PROPOSITION XII. THEOREM. 

If two circumferences touch each other , either externally or 
internally , the distance of their centers must be equal to the 
sum or difference of their radii. 

It is plain that the centers of the circles and the point of 



contact are in the same straight line; for, if possible,, let the 
point of contact, A, be without the straight line CD. # From 
A let fall upon CD, or CD produced, the perpendicular AE, 
and produce it to B, making BE equal to AE. Then, in the 
triangles ACE, BCE, the side AE is equal to EB, CE is com¬ 
mon, and the angle AEC is equal to the angle BEQ; there¬ 
fore AC is equal to CB (Prop. VI., B. I.), and the point B is 
in the circumference ABF. In the same manner, it may be 
shown to be in the circumference ABG, and hence the point 









BOOK III. 


53 


B is in both circumferences. Therefore the two circumfe¬ 
rences have two points, A and B, in common ; that is, they cut 
each other, which is contrary to the hypothesis. Therefore, 
the point of contact can not be without the line joining the 
centers; and hence, when the circles touch each other exter¬ 
nally, the distance of the centers CD is equal to the sum of 
the radii CA, DA; and when they touch internally, the dis¬ 
tance CD is equal to the difference of the radii CA, DA. 
Therefore, if two circumferences, &c. 

Cor. If two circumferences touch each other, externally or 
internally, their centers and the point of contact are in the 
same straight line. 


PROPOSITION XIII. THEOREM. 

If two circumferences cut each other , the distance between 
their centers is less than the sum of their radii , and greater 
than their difference . 

Let two circumferences cut each 
other in the point A. Draw the ra¬ 
dii CA, DA; then, because any two 
sides of a triangle are together great¬ 
er than the third side (Prop. VIII., B. 

I.), CD must be less than the sum of 
AD and AC. Also, DA must be less 
than the sum of CD and CA; or, subtracting CA from these 
unequals (Axiom 5, B. I.), CD must be greater than the dif¬ 
ference between DA and CA. Therefore, if two circumfe¬ 
rences, &c. 



PROPOSITION XIV. THEOREM. 

In equal circles , angles at the center have the same ratio 
with the intercepted arcs. 


Case first. When 
the angles are in the 
ratio of two whole 
numbers. 

Let ABG, DFH 
be equal circles, and 
let the angles ACB, 

DEF at their cen¬ 
ters be in the ratio of two whole numbers; then will 
the angle ACB : angle DEF : : arc AB : arc DF. 





54 


GEOMETRY. 



Suppose, for example, that the angles ACB, DEF are to 
each other as 7 to 4; or, which is the same thing, suppose 
that the angle M, which may serve as a common measure, 
is contained seven times in the angle ACB, and four times in 
the angle DEF. The seven partial angles into which ACB 
is divided, being each equal to any of the four partial angles 
into which DEF is divided, the partial arcs will also be 
equal to each other (Prop. IV.), and the entire arc AB will 
be to the entire arc DF as 7 to 4. Now the same reasoning 
would apply, if in place of 7 and 4 any whole numbers what¬ 
ever were employed ; therefore, if the ratio of the angles 
ACB, DEF can be expressed in whole numbers, the arcs AB, 
DF will be to each other as the angles ACB, DEF. 

Case second. When the ratio of the angles can not be ex¬ 
pressed by whole numbers. 

Let ACB, ACD be two an¬ 
gles having any ratio whatev¬ 
er. Suppose ACD to be the 
smaller angle, and let it be 
placed on tfte greater; then 
will the angle ACB : angle 
ACD : : arc AB : arc AD. 

For, if this proportion is not true, the first three terms re¬ 
maining the same, the fourth must be greater or less than 
AD. Suppose it to be greater, and that we have 

Angle ACB : angle ACD : : arc AB : arc AI. 

Conceive the arc AB to be divided into equal parts, each 
less than DI; there will be at least one point of division be¬ 
tween D and I. Let H be that point, and join CH. The 
arcs AB, AH will be to each other in the ratio of two whole 
numbers, and, by the preceding case, we shall have 

Angle ACB : angle ACH : : arc AB : arc AH. 

Comparing these two proportions with each other, and ob¬ 
serving that the antecedents are the same, we conclude that 
the consequents are proportional (Prop. IV., Cor., B. II.) ; 
therefore, 

Angle ACD : angle ACH : : arc AI : arc AH. 

But the arc AI is greater than the arc AH ; therefore the 
angle ACD is greater than the angle ACH (Def. 2, B. II.), 
that is, a part is greater than the whole, which is absurd. 
Hence the angle ACB can not be to the angle ACD as the 
arc AB to an arc greater than AD. 

In the same manner, it may be proved that the fourth term 
of the proportion can not be less than AD ; therefore, it must 
be AD, and we have the proportion 

Angle ACB : angle ACD : : arc AB : arc AD. 

Cor . 1 . Since the angle at the center of a circle, and the 


BOOK 111. 


55 


arc intercepted by its sides, are so related, that when one is 
increased or diminished, the other is increased or diminished 
in the same ratio, we may take either of these quantities as 
the measure of the other. Henceforth we shall take the arc 
AB to measure the angle ACB. It is important to observe, 
that in the comparison of angles, the arcs which measure 
them must be described with equal radii. 

Cor. 2. In equal circles, sectors are to each other as their 
arcs; for sectors are equal when their angles are equal. 


PROPOSITION XV. THEOREM. 

An inscribed angle is measured by half the arc included be¬ 
tween its sides . 

Let BAD be an angle inscribed in the circle BAD. The 
angle BAD is measured by half the arc BD. 

First . Let C, the center of the circle, 
be within the angle BAD. Draw the di¬ 
ameter AE, also the radii CB, CD. 

Because CA is equal to CB, the angle 
CAB is equal to the angle CBA (Prop. X., 

B. I.); therefore the angles CAB, CBA 
are together double the angle CAB. But 
the angle BCE is equal (Prop. XXVII., B. 

I.) to the angles CAB, CBA; therefore, 
also, the angle BCE is double of the angle BAC. Now the 
angle BCE, being an angle at the center, is measured by the 
arc BE ; hence the angle BAE is measured by the half of 
BE. For the same reason, the angle DAE is measured by 
half the arc DE. Therefore, the whole angle BAD is meas¬ 
ured by half the arc BD. 

Second. Let C, the center of the circle, 
be without the angle BAD. Draw the di¬ 
ameter AE. It may be demonstrated, as 
in the first case, that the angle BAE is 
measured by half the arc BE, and the an¬ 
gle DAE by half the arc DE ; hence their 
difference, BAD, is measured by half of 
BD. Therefore, an inscribed angle, &c. 

Cor. 1 . All the angles BAC, BDC, &c., 
inscribed in the same segment are equal, for they are all 
measured by half the same arc BEC. 

Cor. 2. Every angle inscribed in a semicircle is a right 
angle, because it is measured by half a semicircumference, 
that is, the fourth part of a circumference. 


A 



A 





56 


GEOMETRY. 


' Cor. 3. Every angle inscribed in a 
segment greater than a semicircle is an 
acute angle, for it is measured by half 
an arc less than a semicircumference. 

Every angle inscribed in a segment 
less than a semicircle is an obtuse an¬ 
gle, for it is measured by half an arc 
greater than a semicircumference. 

Cor. 4. The opposite angles of an in¬ 
scribed quadrilateral, ABEC, are together equal to two right 
angles; for the angle BAC is measured by half the arc BEC, 
and the angle BEC is measured by half the arc BAC; there¬ 
fore the two angles BAC, BEC, taken together, are measured 
by half the circumference ; hence their sum is equal to two 
right angles. 



PROPOSITION XVI. THEOREM. 

The angle formed by a tangent and a chord , is measured by 
half the arc included between its sides. 

Let the straight line BE touch the D 

circumference ACDF in the point A, 
and from A let the chord AC be 
drawn ; the angle BAC is measured by ^ 
half the arc AFC. * 

From the point A draw the diameter 
AD. The angle BAD is a right angle 
(Prop. IX.), and is measured by half 
the semicircumference AFD ; also, the 
angle DAC is measured by half the arc DC (Prop. XV.); 
therefore, the sum of the angles BAD, DAC is measured by 
half the entire arc AFDC. 

In the same manner, it may be shown that the angle CAE 
is measured by half the arc AC, included between its sides. 






BOOK IV. 


57 


BOOK IV. 


THE PROPORTIONS OF FIGURES. 


Definitions. 


1. Equal figures are such as, being applied the one to the 
other, coincide throughout.^ Thus, two circles having equal 
radii are equal; and two triangles, having the three sides of 
the one equal to the three sides of the other, each to each, 
are also equal. 

2. Equivalent figures are such as contain equal areas. 
Two figures may be equivalent, however dissimilar. Thus, 
a circle may be equivalent to a square, a triangle to a rec¬ 
tangle, &c. 

3. Similar figures are such as have the angles of the one 
equal to the angles of the other, each to each, and the sides 
about the equal angles proportional. Sides which have the 
same position in the two figures, or which are adjacent to 
equal angles, are called homologous . The equal angles may 
also be called homologous angles. 

Equal figures are always similar, but similar figures may 
be very unequal. 

4. Two sides of one figure are said to be reciprocally pro - 
portional to two sides of another, when one side of the first is 
to one side of the second, as the remaining side of the sec¬ 
ond is to the remaining side of the first. 

5. In different circles, similar arcs , sectors , or segments , are 
those which correspond to equal angles at the center. 


Thus, if the angles A and D are 
equal, the arc BC will be similar to 
the arc EF, the sector ABC to the 
sector DEF, and the segment BGC 
to the segment EHF. 



6. The altitude of a triangle is the perpen 
dicular let fall from the vertex of an angle 
on the opposite side, taken as a base. 







58 


geometry. 


7. The altitude of a parallelogram is the 
distance between two parallel sides. 


8. The altitude of a trapezoid is the distance 
between its parallel sides. 


A 


PROPOSITION I. THEOREM. 

Parallelograms which have equal bases and equal altitudes , 
are equivalent. 

Let the parallelo¬ 
grams ABCD, ABEF 
be placed so that their 
equal bases shall coin¬ 
cide with each other. 

Let AB be the common 
base ; and, since the two parallelograms are supposed to have 
the same altitude, their upper bases, DC, FE, will be in the 
same straight line parallel to AB. 

Now, because ABCD is a parallelogram, DC is equal to 
AB (Prop. XXIX., B. I.). For the same reason, FE is equal 
to AB, wherefore DC is equal to FE; hence, if DC and FE 
be taken away from the same line DE, the remainders CE 
and DF will be equal. But AD is also equal to BC, and AF 
to BE ; therefore the triangles DAF, CBE are mutually equi¬ 
lateral, and consequently equal. 

Now if from the quadrilateral ABED we take the triangle 
ADF, there will remain the parallelogram ABEF; and if 
from the same quadrilateral we take the triangle BCE, there 
will remain the parallelogram ABCD. Therefore, the two 
parallelograms ABCD, ABEF, which have the same base 
and the same altitude, are equivalent. 

Cor. Every parallelogram is equivalent to the rectangle 
which has the same base and the same altitude. 



PROPOSITION II. THEOREM. 

Every triangle is half of the parallelogram which has the 
same base and the same altitude. 

Let the parallelogram ABDE and the triangle ABC have 
the same base, AB, and the same altitude; the triangle is 
half of the parallelogram. 












BOOK IV. 


59 


Complete the parallelogram ABFC ; q 
then the parallelogram ABFC is equiv¬ 
alent to the parallelogram ABDE, be¬ 
cause they have the same base and the 
same altitude (Prop. I.). But the tri¬ 
angle ABC is half of the parallelogram 
ABFC (Prop. XXIX., Cor., B. I.) ; wherefore the triangle 
ABC is also half of the parallelogram ABDE. Therefore, 
every triangle, &c. 

Cor. 1. Every triangle is half of the rectangle which has 
the same base and altitude. 

Cor. 2. Triangles which have equal bases and equal alti¬ 
tudes are equivalent. 



PROPOSITION III. THEOREM. 


Two rectangles of the same altitude , are to each other as their 
bases. 


D 


E 


13 


Let ABCD, AEFD be two rec¬ 
tangles which have the common al¬ 
titude AD ; they are to each other 
as their bases AB, AE. 

Case first. When the bases are in 
the ratio of two whole numbers, for 
example, as 7 to 4. If AB be divided into seven equal parts, 
AE will contain four of those parts. At each point of divis¬ 
ion, erect a perpendicular to the base ; seven partial rectan¬ 
gles will thus be formed, all equal to each other, since they 
have equal bases and altitudes (Prop. I.). The rectangle 
ABCD will contain seven partial rectangles, while AEFD 
will contain four; therefore the rectangle ABCD is to the 
rectangle AEFD as 7 to 4, or as AB to AE. The same rea¬ 
soning is applicable to any other ratio than that of 7 to 4; 
therefore, whenever the ratio of the bases can be expressed 
in whole numbers, we shall have 

ABCD : AEFD : : AB : AE. 

Case second. When the ratio of the bases can not be ex¬ 
pressed in whole numbers, it is still true that 
ABCD : AEFD : : AB : AE. 

For, if this proportion is not true, the 
first three terms remaining the same, the 
fourth must be greater or less than AE. 

Suppose it to be greater, and that we have 
ABCD : AEFD : : AB : AG. 

Conceive the .line AB to be divided into 



EHG- B 













GO 


GEOMETRY. 


equal parts, each less than EG; there will 
be at least one point of division between 
E and G. Let H be that point, and draw 
the perpendicular HI. The bases AB, AH 
will be to each other in the ratio of two 
whole numbers, and by the preceding case 
we shall have 

ABCD : AHID : : AB : AH. 

But, by hypothesis, we have 

ABCD : AEFD : : AB : AG. 

In these two proportions the antecedents are equal; there¬ 
fore the consequents are proportional (Prop. IV., Cor., B. II.), 
and we have 

AHID : AEFD : : AH : AG. 

But AG is greater than AH ; therefore the rectangle 
AEFD is greater than AHID (Def. 2, B. II.) ; that is, a part is 
greater than the whole, which is absurd. Therefore ABCD 
can not be to AEFD as AB to a line greater than AE. 

In the same manner, it may be shown that the fourth term 
of the proportion can not be less than AE; hence it must be 
AE, and we have the proportion 

ABCD : AEFD : : AB : AE. 

Therefore, two rectangles, &c. 



PROPOSITION IV. THEOREM. 

Any two rectangles are to each other as the products of their 
bases by their altitudes . 

Let ABCD, AEGF be two rectangles ; the ratio of the rec¬ 
tangle ABCD to the rectangle AEGF, is the same with the 
ratio of the product of AB by AD, to the product of AE by 
AF; that is, 

ABCD : AEGF : : AB X AD : AE x AF. 

Having placed the two rectangles so 
that the angles at A are vertical, pro- ] 
duce the sides GE, CD till they meet in 
H. The two rectangles ABCD, AEHD - 
have the same altitude AD; they are, 
therefore, as their bases AB, AE (Prop. 

III.). So, also, the rectangles AEHD, _ 

AEGF, having the same altitude AE, G ^ 

are to each other as their bases AD, AF. Thus, we have the 

two proportions 

ABCD : AEHD : : AB : AE, 

AEHD : AEGF : : AD: AF. 











BOOK IV. 


61 


Hence (Prop. XI., Cor., B. II.), 

ABCD : AEGF : : AB X AD : AE X AF. 

Scholium. Hence we may take as the measure of a rec¬ 
tangle, the product of its base by its altitude; provided we un¬ 
derstand by it the product of two numbers, one of which is 
the number of linear units contained in the base, and the oth¬ 
er the number of linear units contained in the altitude. 


PROPOSITION V. THEOREM. 


The area of a parallelogram is equal to the product of its 
base by its altitude. 

Let ABCD be a parallelogram, AF its 
altitude, and AB its base; then is its sur¬ 
face measured by the product of AB by 
AF. For, upon the base AB, construct a 
rectangle having the altitude AF ; the par¬ 
allelogram ABCD is equivalent to the rec¬ 
tangle ABEF (Prop. I., Cor.). But the rectangle ABEF is 
measured by AB X AF (Prop. IV., Schol.); therefore the area 
of the parallelogram ABCD is equal to AB X AF. 

Cor. Parallelograms of the same base are to each other as 
their altitudes, and parallelograms of the same altitude are 
to each other as their bases; for magnitudes have the same 
ratio that their equimultiples have (Prop. VIII., B. II.). 



PROPOSITION VI. THEOREM. 

The area of a triangle is equal to half the product of its 
base by its altitude. 

Let ABC be any triangle, BC its base, and 
AD its altitude ; the area of the triangle ABC 
is measured by half the product of BC by AD. 

For, complete the parallelogram ABCE. 

The triangle ABC is half of the parallelo- _ 
gram ABCE (Prop. II.); but the area of the 
parallelogram is equal to BC X AD (Prop. V.); hence the 
area of the triangle is equal to one half of the product of 
BC by AD. Therefore, the area of a triangle, &c. 

Cor. 1 . Triangles of the same altitude are to each other 
as their bases, and triangles of the same base are to each oth¬ 
er as their altitudes. 

Cor. 2. Equivalent triangles, whose bases are equal, have 









62 


GEOMETRY. 


equal altitudes; and equivalent triangles, whose altitudes are 
equal, have equal bases. # 


PROPOSITION VII. THEOREM. 

The area of a trapezoid is equal to half the product of its 
altitude by the sum of its parallel sides. 

Let ABCD be a trapezoid, DE its al- 33 
titude, AB and CD its parallel sides; JT 
its area is measured by half the product / 
of DE, by the sum of its sides AB, CD,. y " " 

Bisect BC in F, and through F draw / 

GH parallel to AD, and produce DC to a e 
H. In the two triangles BFG, CFH, 
the side BF is equal to CF by construction, the vertical an¬ 
gles BFG, CFH are equal (Prop. V., B. I.), and the angle 
FCH is equal to the alternate angle FBG, because CH and 
BG are parallel (Prop. XXIII., B. I.); therefore the triangle 
CFH is equal to the triangle BFG. Now, if from the whole 
figure, ABFHD, we take away the triangle CFH, there will 
remain the trapezoid ABCD; and if from the same figure, 
ABFHD, we take away the equal triangle BFG, there will 
remain the parallelogram AGHD. Therefore the trapezoid 
ABCD is equivalent to the parallelogram AGHD, and is 
measured by the product of AG by DE. 

Also, because AG is equal to DH, and BG to CH, there¬ 
fore the sum of AB and CD is equal to the sum of AG and 
DH, or twice AG. Hence AG is equal to half the sum of 
the parallel sides AB, CD; therefore the area of the trape¬ 
zoid ABCD is equal to half the product of the altitude DE 
by the sum of the bases AB, CD. 

Cor. If through the point F, the middle of BC, we draw 
FK parallel to the base AB, the point K will also be the mid¬ 
dle of AD. For the figure AKFG is a parallelogram, as 
also DKFH, the opposite sides being parallel. Therefore 
AK is equal to FG, and DK to HF. But FG is equal to FH, 
since the triangles BFG, CFH are equal; therefore AK is 
equal to DK. s 

Now, since KF is equal to AG, the area of the trapezoid is 
equal to DE X KF. Hence the area of a trapezoid is equal to 
its altitude, multiplied by the line which joins the middle points 
of the sides which are not parallel . 








BOOK IV. 


63 


PROPOSITION VIII. THEOREM. 


If a straight line is divided into any two parts , the square of 
the whole line is equivalent to the squares of the two parts y to¬ 
gether with twice the rectangle contained by the parts . 

Let the straight line AB be divided into any two parts in 
C; the square on AB is equivalent to the squares on AC, 
CB, together with twice the rectangle contained by AC, CB; 
that is, 

AB 2 , or (AC+CB) 3 =AC 3 +CB 2 +2ACxCB. 

Upon AB describe the square ABDE; E h D 

take AF equal to AC, through F draw FG 
parallel to AB, and through C draw CH par- F * \ & 

allel to AE. 

The square ABDE is divided into four 

parts: the first, ACIF, is the square on AC,_ 

since AF was taken equal to AC. The sec- A c . 15 

ond part, IGDH, is the square on CB; for, because AB is 
equal to AE, and AC to AF, therefore BC is equal to EF 
(Axiom 3, B. I.). But, because BCIG is a parallelogram, 
GI is equal to BC ; and because DEFG is a parallelogram, 
DG is equal to EF (Prop. XXIX., B. I.); therefore HIGD is 
equal to a square described on BC. If these two parts are 
taken from the entire square, there will remain the two rect¬ 
angles BCIG, EFIH, each of which is measured by AC X 
CB; therefore the whole square on AB is equivalent to the 
squares on AC and CB, together with twice the rectangle of 
AC xCB. Therefore, if a straight line, &c. 

Cor. The square of any line is equivalent to four times the 
square of half that line. For, if AC is equal to CB, the four 
figures AI, CG, FH, ID become equal squares. 

Scholium. This proposition is expressed algebraically 
thus: 

(a+6) 3 =a a +2a6+6 2 . 


PROPOSITION IX. THEOREM. 

The square described on the difference of two lines , is equiv¬ 
alent to the sum of the squares of the lines , diminished by twice 
the rectangle contained by the lines. 

Let AB, BC be any two lines, and AC their difference; 
the square described on AC is equivalent to the sum of the 





64 


GEOMETRY. 


B 


squares on AB and CB, diminished by twice the rectangle 
contained by AB, CB ; that is, 

AC 2 , or (AB — BC) 2 =AB 2 +BC 2 —2ABxBC. 

Upon AB describe the square ABKF; j, f G K 
take AE equal to AC, through C draw 

CG parallel to BK, and through E draw _}- |h 

HI parallel to AB, and complete the 1 E 
square EFLI. 

Because AB is equal to AF, and AC to 
AE; therefore CB is equal to EF, and GK 
to LF. Therefore LG is equal to FK or AB; and hence the 
two rectangles CBKG, GLID are each measured by AB X 
BC. If these rectangles are taken from the entire figure 
ABKLIE, which is equivalent to AB 2 +BC 2 , there will evi¬ 
dently remain the square ACDE. Therefore, the square 
described, &c. 

Scholium . This proposition is expressed algebraically 
thus: 

( a—&) 2 =« 2 —2 ab+b\ 

Cqr. (a+by~-(a — by=4ab. 


PROPOSITION X. THEOREM. 


The rectangle contained by the sum and difference of two 
lines , is equivalent to the difference of the squares of those lines. 


Let AB, BC be any two lines ; the rectangle contained by 
the sum and difference of AB and BC, is equivalent to the 
difference of the squares on AB and BC; that is, 

(AB+BC) x (AB —BC) = AB 2 —BC 2 . 

Upon AB describe the square ABKF, 
and upon AC describe the square ACDE; 
produce AB so that BI shall be equal to 
BC, and complete the rectangle AILE. 

The base AI of the rectangle AILE is 
the sum of the two lines AB, BC, and its 
altitude AE is the difference of the same 


G K 




H j 

D 




C B I 

lines; therefore AILE is the rectangle contained by the sum 
and difference of the lines AB, BC. But this rectangle is 
composed of the two parts ABHE and BILH ; and the part 
BILH is equal to the rectangle EDGF, for BH is equal to 
DE, and BI is equal to EF. Therefore AILE is equivalent 
to the figure ABHDGF. But ABHDGF is the excess of the 
square ABKF above the square DHKG, which is the square 
ofBC; therefore, 

(AB+BC) x(AB—BC)=AB 2 —BC 2 . 
















BOOK IV. 


65 


Scholium. This proposition is expressed algebraically 
thus: 

(«+&) X (a — b)=a 2 — b 2 . 


PROPOSITION XI. THEOREM. 

In any right-angled triangle , the square described on the hy - 
pothenuse is equivalent to the sum of the squares on the other 
two sides. 

Let ABC be a right-angled triangle, 
having the right angle BAG; the 
square described upon the side BC is 
equivalent to the sum of the squares 
upon BA, AC. 

On BC describe the square BCED, 
and on BA, AC the squares BG, CH; 
and through A draw AL parallel to 
BD, and join AD, FC. 

Then, because each of the angles 
BAC, BAG is a right angle, CA is in 
the same straight line with AG (Prop. III., B. I.). For the 
same reason, BA and AH are in the same straight line. 

The angle ABD is composed of the angle ABC and the 
right angle CBD. The angle FBC is composed of the same 
angle ABC and the right angle ABF; therefore the whole 
angle ABD is equal to the angle FBC. But AB is equal to 
BF, being sides of the same square; and BD is equal to BC 
for the same reason ; therefore the triangles ABD, FBC have 
two sides and the included angle equal; they are therefore 
equal (Prop. VI., B. I.). 

But the rectangle BDLK is double of the triangle ABD, 
because they have the same base, BD, and the same altitude, 
BK (Prop. II., Cor. 1); and the square AF is double of the 
triangle FBC, for they have the same base, BF, and the same 
altitude, AB. Now the doubles of equals are equal to one 
another (Axiom 6, B. I.); therefore the rectangle BDLK is 
equivalent to the square AF. 

In the same manner, it may be demonstrated that the rec¬ 
tangle CELK is equivalent to the square AI; therefore the 
whole square BCED, described on the hypothenuse, is equiv¬ 
alent to the two squares ABFG, ACIH, described on the two 
other sides ; that is, 

BC 2 =AB 2 +AC 2 . 

Cor. 1 . The square of one of the sides of a right-angled 
E 




66 


GEOMETRY. 


triangle, is equivalent to the square of the hypothenuse, dimin¬ 
ished by the square of the other side ; that is, 

AB 2 = BC 2 — AC 2 . 

Cor. 2. The square BCED, and the rectangle BKLD, hav¬ 
ing the same altitude, are to each other as their bases BC, 
BK (Prop. III.). But the rectangle BKLD is equivalent to 
the square AF; therefore, 

BC 2 : AB 2 : : BC : BK. 

In the same manner, 

BC 2 : AC 2 : : BC : KC. 

Therefore (Prop. IV., Coi\, B. II.), 

AB 2 : AC 2 : : BK : KC. 

That is, in any right-angled triangle, if a line be drawn 
from the right angle perpendicular to the hypothenuse, the 
squares of the two sides are proportional to the adjacent seg¬ 
ments of the hypothenuse ; also, the square of the hypothenuse 
is to the square of either of the sides , as the hypothenuse is to 
the segment adjacent to that side. 

Cor. 3. Let ABCD be a square, and AC its 
diagonal; the triangle ABC being right-angled 
and isosceles, we have 

AC 2 =AB 2 +BC 2 =2AB 2 ; 
therefore the square described on the diagonal of a 
square , is double of the square described on a side. 

If we extract the square root of each mem¬ 
ber of this equation, we shall have 

AC=AB v /2; or AC : AB : : </2 : 1. 



PROPOSITION Xir. THEOREM. 


In any triangle , the square of a side opposite an acute angle , 
is less than the squares of the base and of the other side , by 
twice the rectangle contained by the base , and the distance from 
the acute angle to the foot of the perpendicular let fall from the 
opposite angle. 


Let ABC be any triangle, and the angle at C one of its 
acute angles, and upon BC let fall the perpendicular AD from 
the opposite angle ; then will 

AB 2 =BC 2 +AC 2 —2BC xCD. 


First. When the perpendicular falls with¬ 
in the triangle ABC, we have BD = BC — CD, 
and therefore BD 2 =BC 2 +CD 2 —2BC xCD 
(Prop. IX.). To each of these equals add 
AD 2 ; then BD 2 -f AD 2 =BC 2 4-CD 2 -f AD 2 — 
2BC xCD. But in the right-angled triangle 






BOOK IV. 


67 


ABD, BD 2 +AD 2 =AB 2 ; and in the triangle ADC, CD’-f 
AD 2 —AC 2 (Prop. XI.); therefore 

AB 2 =BC 2 +AC 2 — 2 BC x CD. 

Secondly. When the perpendicular falls a 
without the triangle ABC, we have BD = 

CD—BC, and therefore BD 2 =CD 2 +BC 2 — 

2CDxBC (Prop. IX.). To each of these 
equals add AD 2 ; then BD 2 -f AD 2 =CD 2 -f AD 2 
4-BC 2 —2CD xBC. But BD 2 +AD 2 =AB 2 ; 
and CD 2 +AD 2 =AC 2 ; therefore D B c 

AB 2 =BC 2 +AC 2 —2BC xCD. 

Scholium. When the perpendicular AD falls upon AB, 
this proposition reduces to the same as Prop. XI., Cor. 1 . 



PROPOSITION XIII. THEOREM. 

In obtuse-angled triangles , the square of the side opposite 
the obtuse angle , is greater than the squares of the base and the 
other side , by twice the rectangle contained by the base , and the 
distance from the obtuse angle, to the foot of the perpendicular 
let fall from the opposite angle on the base produced. 

Let ABC be an obtuse-angled triangle, having the obtuse 
angle ABC, and from the point A let AD be drawn perpen¬ 
dicular to BC produced ; the square of AC is greater than 
the squares of AB, BC by twice the rectangle BCxBD. 

For CD is equal to BC+BD ; therefore CD 2 a 
=BC 2 +BD 2 + 2 BCxBD (Prop. VIII.). To 
each of these equals add AD 2 ; then CD 2 -f 
AD 2 = BC 2 +BD 2 +AD 2 + 2 BCxBD. But AC 2 
is equal to CD 2 +AD 2 (Prop. XI.), and AB 2 is 
equal to BD 2 +AD 2 ; therefore AC 2 =BC 2 + 
AB 2 + 2 BCxBD. Therefore, in obtuse-an¬ 
gled triangles, &c. 

Scholium. The right-angled triangle is the only one in 
which the sum of the squares of two sides is equivalent to the 
square on the third side; for, if the angle contained by the 
two sides is acute, the sum of their squares is greater than 
the square of the opposite side; if obtuse, it is less. 



PROPOSITION XIV. THEOREM. 

In any triangle , if a straight line is drawn from the vertex 
to the middle of the base , the sum of the squares of the other two 
sides is equivalent to twice the square of the bisecting line, to¬ 
gether with twice the square of half the base. 

Let ABC be a triangle having a line AD drawn from the 






68 


GEOMETRY. 


middle of the base to the opposite angle; the squares of BA 
and AC are together double of the squares of AD and BD. 

From A draw AE perpendicular to BC; 
then, in the triangle ABD, by Prop. XIII., 
AB a =AD a +DB a +2DBxDE; 
and, in the triangle ADC, by Prop. XII., 

AC a =AD a +DC a —2DC X DE. 

Hence, by adding these equals, and ob¬ 
serving that BD=DC, and therefore BD 2 = 

DC 3 , and DB XDE=DC xDE, we obtain 
AB a +AC a =2AD a +2DB a 
Therefore, in any triangle, &c. 



PROPOSITION XV. THEOREM. 

In every parallelogram the squares of the sides are together 
equivalent to the squares of the diagonals. 



AB a +BC a +CD a +AD a =4BE a +4AE a . 


But 4BE a =BD a ,and 4AE a =AC a (Prop. VIII., Cor.); there¬ 
fore 

AB a +BC a +CD a +AD a =BD a +AC a . 

Therefore, in every parallelogram, &c. 


PROPOSITION XVI. THEOREM. 

If a straight line he drawn parallel to the base of a triangle , 
it will cut the other sides proportionally; and if the sides be 
cut proportionally , the cutting line will be parallel to the base 
of the triangle. 

Let DE be drawn parallel to BC, the base of the triangle 
ABC; then will AD : DB :: AE : EC. 





BOOK IV. 


69 


Join BE and DC ; then the triangle BDE is a 

equivalent to the triangle DEC, because they A 

have the same base, DE, and the same altitude, / \ 

since their vertices B and C are in a line par- / \ 

allel to the base (Prop. II., Cor. 2). D /_\ E 

The triangles ADE, BDE, whose common \ 

vertex is E, having the same altitude, are to \\ 

each other as their bases AD, DB (Prop. VI., B c 

Cor. 1); hence 

ADE : BDE : : AD : DB. 

The triangles ADE, DEC, whose common vertex is D, 
having the same altitude, are to each other as their bases 
AE, EC; therefore 

ADEDEC : : AE : EC. 

But, since the triangle BDE is equivalent to the triangle 
DEC, therefore (Prop. IV., B. II.), 

AD : DB : : AE : EC. 

Conversely, let DE cut the sides AB, AC, so that AD : DB 
: : AE : EC; then DE will be parallel to BC. 

For AD : DB : : ADE : BDE (Prop. VI., Cor. 1); and AE 
: EC : : ADE : DEC; therefore (Prop. IV., B. II.), ADE : 
BDE : : ADE : DEC; that is, the triangles BDE, DEC have 
the same ratio to the triangle ADE; consequently, the trian¬ 
gles BDE, DEC are equivalent, and having the same base DE, 
their altitudes are equal (Prop. VI., Cor. 2), that is, they are 
between the same parallels; Therefore, if a straight line, &c. 

'Cor. 1. Since, by this proposition, AD : DB; ; AE : EC; 
by composition, AD+DB : AD : : AE+EC : AE (Prop. VI., 
B. II.), or AB : AD : : AC : AE; also, AB : BD : : AC : EC. 


Cor. 2. If two lines be drawn parallel to the a 

base of a triangle, they will divide the other sides A 

proportionally. For, because FG is drawn J \ 
parallel to BC, by the preceding proposition, y \ E 

AF : FB : : AG : GC. Also, by the last cor- j?f _Vq- 

ollary, because DE is parallel to FG, AF : DF f y 


(Prop. IV., Cor., B. II.). Aiso, AD : DF : : 13 C 


AE : EG. 

Cor. 3. If any number of lines be drawn parallel to the 
base of a triangle, the sides will be cut proportionally. 


PROPOSITION XVII. THEOREM. 

The line which bisects the vertical angle of a triangle, di¬ 
vides the base into two segments, which are proportional to the 
adjacent sides. 






70 


GEOMETRY. 


Let the angle BAC of the triangle ABC be bisected by tht 
straight line AD ; then will E 

BD : DC : : BA : AC. v " 

Through the point B draw BE par¬ 
allel to DA, meeting CA produced in E. 

The triangle ABE is isosceles. For, 
since AD is parallel to EB, the angle 
ABE is equal to the alternate angle 
DAB (Prop. XXIII., B. I.); and the exterior angle CAD is 
equal to the interior and opposite angle AEB. But, by hy¬ 
pothesis, the angle DAB is equal to the angle DAC ; there¬ 
fore the angle ABE is equal to AEB, and the side AE to the 
side AB (Prop. XI., B. I.). 

And because AD is drawn parallel to BE, the base of the 
triangle BCE (Prop. XVI.), 

BD : DC : : EA : AC. 

But AE is equal to AB, therefore 

BD : DC : : BA : AC. 

Therefore, the line, &c. 

Scholium. The line which bisects the exterior angle of a 
triangle, divides the base produced into segments, which are 
proportional to the adjacent sides. 

Let the line AD bisect the exterior 
angle CAE of the triangle ABC ; then 
BD : DC : : BA : AC. 

Through C draw CF parallel to 
AD; then it may be proved, as in the r 

preceding proposition, that the angle C 

ACF is equal to the angle AFC, and AF equal to AC. And 
because FC is parallel to AD (Prop. XVI., Cor. I), BD : DC 
: : BA : AF; but AF is equal to AC; therefore 
BD : DC : : BA : AC. 




PROPOSITION XVIII. THEOREM. 


Equiangular triangles have their homologous sides propor¬ 
tional r , and are similar. 


Let ABC, DCE be two equiangular 
triangles, having the angle BAC equal to 
the angle CDE, and the angle ABC equal 
to the angle DCE, and, consequently, the 
angle ACB equal to the angle DEC ; then 
the homologous sides will be proportion¬ 
al, and we shall have 

BC : CE : : BA : CD : : AC 






BOOK IV. 


71 


Place the triangle DCE so that the side CE may be con¬ 
tiguous to BC, and in the same straight line with it; and pro¬ 
duce the sides BA, ED till they meet in F. 

Because BCE is a straight line, and the angle ACB is 
equal to the angle DEC, AC is parallel to EF (Prop. XXII., 
B. I.). Again, because the angle ABC is equal to the angle 
DCE, the line AB is parallel to DC; therefore the figure 
ACDF is a parallelogram, and, consequently, AF is equal to 
CD, and AC to FD (Prop. XXIX., B. I.). 

And because AC is parallel to FE, one of the sides of the 
triangle FBE, BC : CE : : BA : AF (Prop. XVI.) ; but AF is 
equal to CD ; therefore 

BC : CE : : BA : CD. 

Again, because CD is parallel to BF, BC : CE : : FD : DE. 
But FD is equal to AC ; therefore 

BC : CE : : AC : DE. 

And, since these two proportions contain the same ratio 
BC : CE, we conclude (Prop. IV., B. II.) 

BA : CD : : AC : DE. 

Therefore the equiangular triangles ABC, DCE have their 
homologous sides proportional; hence, by Def. 3, they are 
similar. 

Cor. Two triangles are similar when they have two an¬ 
gles equal, each to each, for then the third angles must also 
be equal. 

Scholium. In similar triangles the homologous sides are 
opposite to the equal angles; thus, the angle ACB being 
equal to the angle DEC, the side AB is homologous to DC, 
and so with the other sides. 


PROPOSITION XIX. THEOREM. 


Two triangles which have their homologous sides proportion - 
al , are equiangular and similar. 

Let the triangles ABC, DEF 
have their sides proportional, so 
that BC : EF : : AB : DE : : AC 
. DF; then will the triangles 
have their angles equal, viz.: 
the angle A equal to the angle 
D, B equal to E, and C equal to 
F. 

At the point E, in the straight 
line EF, make the angle FEG equal to B, and at the point F 
make the angle EFG equal to C ; the third angle G will be 





72 


GEOMETRY. 


equal to the third angle A, and the two triangles ABC, GEF 
will be equiangular (Prop. XXVII., Cor. 2, B. I.); therefore, 
by the preceding theorem, 

BC : EF : : AB : GE. 

But, by hypothesis, 

BC:EF::AB:DE; 
therefore GE is equal to DE. 

Also, by the preceding theorem, 

BC : EF : : AC : GF; 

but, by hypothesis, 

BC : EF : : AC : DF; 

consequently, GF is equal to DF. Therefore the triangles 
GEF, DEF have their three sides equal, each to each ; hence 
their angles also are equal (Prop. XV., B. I.). But, by con¬ 
struction, the triangle GEF is equiangular to the triangle 
ABC ; therefore, also, the triangles DEF, ABC are equiangu¬ 
lar and similar. Wherefore, two triangles, &c. 


PROPOSITION XX. THEOREM. 

Two triangles are similar , when they have an angle of the 
one equal to an angle of the other , and the sides containing 
those angles proportional. 

Let the triangles ABC, DEF have the angle A of the one, 
equal to the angle D of the other, and let AB : DE : : AC : 
DF; the triangle ABC is similar to the triangle DEF. 

Take AG equal to DE, also AH a 
equal to DF, and join GH. Then K 
the triangles AGH, DEF are equal, / \ 
since two sides and the included / \ 

angle in the one, are respectively q 1 _\h 

equal to two sides and the included / \ 

angle in the other (Prop. VI., B. I.). /_\ 

But, by hypothesis, AB : DE : : AC B C 

: DF; therefore 

AB : AG : : AC : AH; 
that is, the sides AB, AC, of the triangle ABC, are cut pro¬ 
portionally by the line GH; therefore GH is parallel to BC 
(Prop. XVI.); hence (Prop. XXIII., B. I.) the angle AGH is 
equal to ABC, and the triangle AGH is similar to the trian¬ 
gle ABC. But the triangle DEF has been shown to be 
equal to the triangle AGH; hence the triangle DEF is simi¬ 
lar to the triangle ABC. Therefore, two triangles, &c. 





BOOK IV. 


73 


PROPOSITION XXI. THEOREM. 

Two triangles are similar , when they have their homologous 
sides parallel or perpendicular to each other. 

Let the triangles ABC, abc , DEF have their homologous 
sides parallel or perpendicular to each other; the triangles 
are similar. 

First . Let the homologous 
sides be parallel to each other. 

If the side AB is parallel to 
ab , and BC to be, the angle B 
is equal to the angle b (Prop. 

XXVI., B. I.); also, if AC is 
parallel to ac , the angle C is 
equal to the angle c; and hence 
the angle A is equal to the 
angle a. Therefore the trian¬ 
gles ABC, abc are equiangular, and consequently similar. 

Secondly. Let the homologous sides be perpendicular to 
each other. Let the side DE be perpendicular to AB, and 
the side DF to AC. Produce DE to I, and DF to H; then, 
in the quadrilateral AIDH, the two angles I and H are right 
angles. But the four angles of a quadrilateral are together 
equal to four right angles (Prop. XXVIII., Cor. 1, B. I.); 
therefore the two remaining angles IAH, IDH are together 
equal to two right angles. But the two angles EDF, IDH 
are together equal to two right angles (Prop. II., B. I.); 
therefore the angle EDF is equal to IAH or BxiC. 

In the same manner, if the side EF is also perpendicular to 
BC, it may be proved that the angle DFE is equal to C, and, 
consequently, the angle DEF is equal to B; hence the trian¬ 
gles ABC, DEF are equiangular and similar. Therefore, two 
triangles &c. 

Scholium. When the sides of the two triangles are paral¬ 
lel, the parallel sides are homologous ; but when the sides are 
perpendicular to each other, the perpendicular sides are ho¬ 
mologous. Thus DE is homologous to AB, DF to AC, and 
EF to BC. 






74 


GEOMETRY. 


PROPOSITION XXII. THEOREM. 


In a right-angled triangle , if a perpendicular is drawn from 
the right angle to the hypothenuse ; 

1 st. The triangles on each side of the perpendicular are sim¬ 
ilar to the whole triangle and to each other. 

2d. The perpendicular is a mean proportional between the 
segments of the hypothenuse. 

3d. Each of the sides is a mean proportional between the hy¬ 
pothenuse and its segment adjacent to that side. 


Let ABC be a right-angled triangle, hay- a 

ing the right angle BAC, and from the angle 
A let AD be drawn perpendicular to the \ 

hypothenuse BC. _ \ 

First. The triangles ABD, ACD are sim- B DC 

ilar to the whole triangle ABC, and to each other. 

The triangles BAD, BAC have the common angle B, also 
the angle BAC equal to BDA, each of them being a right an¬ 
gle, and, therefore, the remaining angle ACB is equal to the 
remaining angle BAD (Prop. XXVII., Cor. 2, B. I.); therefore 
the triangles ABC, ABD are equiangular and similar. In 
like manner, it may be proved that the triangle ADC is equi¬ 
angular and similar to the triangle ABC ; therefore the three 
triangles ABC, ABD, ACD are equiangular and similar to 
each other. 

Secondly. The perpendicular AD is a mean proportional be¬ 
tween the segments BD, DC of the hypothenuse. For, since 
the triangle ABD is similar to the triangle ADC, their ho¬ 
mologous sides are proportional (Def. 3), and we have 
BD : AD : : AD : DC. 

Thirdly. Each of the sides AB, AC is a mean proportional 
between the hypothenuse and the segment adjacent to that 
side. For, since the triangle BAD is similar to the triangle 
BAC, we have 

BC : BA : : BA : BD. 

And, since the triangle ABC is similar to the triangle ACD, 
we have 


BC : CA : : CA : CD. 
Therefore, in a right-angled triangle, &c. 

Cor. If from a point A, in the circumfe¬ 
rence of a circle, two chords AB, AC are 
drawn to the extremities of the diameter 
BC, the triangle BAC will be right-angled 
at A (Prop. XV., Cor. 2, B. III.) ; therefore 



D C 






BOOK IV. 


75 


the perpendicular AD is a mean proportional between BD 
and DC, the two segments of the diameter; that is, 
AD 2 =BDxDC. 


PROPOSITION XXIII. THEOREM. 

Two triangles , having an angle in the one equal to an angle 
in the other , are to each other as the rectangles of the sides 
which contain the equal angles. 

Let the two triangles ABC, ADE have 
the angle A in common; then will the trian¬ 
gle ABC be to the triangle ADE as'the rect¬ 
angle AB X AC is to the rectangle AD X AE. 

Join BE. Then the two triangles ABE, 

ADE, having the common vertex E, have 
the same altitude, and are to each other as 
their bases AB, AD (Prop. VI., Cor. 1); 
therefore 

ABE : ADE : : AB : AD. 

Also, the two triangles ABC, ABE, having the common 
vertex B, have the same altitude, and are to each other as 
their bases AC, AE ; therefore 

ABC : ABE : : AC : AE. 

Hence (Prop. XI., Cor., B. II.). 

ABC : ADE : : AB X AC : AD X AE. 

Therefore, two triangles, &c. 

Cor. 1. If the rectangles of the sides containing the equal 
angles are equivalent, the triangles will be equivalent. 

Cor. 2. Equiangular parallelograms are to each other as 
the rectangles of the sides which contain the equal angles. 



PROPOSITION XXIV. THEOREM. 


Similar triangles are to each other as the squares described 
on their homologous sides. 

Let ABC, DEF be two simi¬ 
lar triangles, having the angle A 
equal to D, the angle B equal to 
E, and C equal to F; then the 
triangle ABC is to the triangle 
DEF as the square on BC is to 
the square on EF. 

By similar triangles, we have (Def. 3) 

AB : DE : : BC : EF. 

Also, BC : EF :: BC : EF. 






76 


GEOMETRY. 


Multiplying together the corresponding terms of these pro¬ 
portions, we obtain (Prop. XI., B. II.), 

ABxBC : DExEF : : BC 3 : EF 3 . 


But, by Prop. XXIII., 

ABC : DEF : : ABxBC : DE XEF; 
hence (Prop. IV., B. II.) 

ABC : DEF : : BC 3 : EF 3 . 
Therefore, similar triangles, &c. 


PROPOSITION XXV. THEOREM. 

Two similar polygons may be divided into the same number 
of triangles , similar each to each , and similarly situated. 

Let ABCDE, FGHIK JC 

be two similar polygons ; 
they may be divided into B 
the same number of sim¬ 
ilar triangles. Join AC, 

AD, FH, FI. A 

Because the polygon 
ABCDE is similar to the E 

polygon FGHIK, the angle B is equal to the angle G (Def. 
3), and AB : BC : : FG : GH. And, because the triangles 
ABC, FGH have an angle in the one equal to an angle in 
the other, and the sides about these equal angles proportion¬ 
al, they are similar (Prop. XX.); therefore the angle BCA 
is equal to the angle GHF. Also, because the polygons are 
similar, the whole angle BCD is equal (Def. 3) to the whole 
angle GHI; therefore, the remaining angle ACD is equal to 
the remaining angle FHI. Now, because the triangles ABC, 
FGH are similar, 

AC : FH : : BC : GH. 

And, because the polygons are similar (Def. 3), 

BC : GH : : CD : HI; 
whence AC : FH : : CD : HI; 

that is, the sides about the equal angles ACD, FHI are pro¬ 
portional ; therefore the triangle ACD is similar to the trian¬ 
gle FHI (Prop. XX.). For the same reason, the triangle 
ADE is similar to the triangle FIK; therefore the similar 
polygons ABCDE, FGHIK are divided into the same num¬ 
ber of triangles, which are similar, each to each, and similar¬ 
ly situated. 

Cor. Conversely, if two polygons are composed of the same 
number of triangles , similar and similarly situated , the poly¬ 
gons are similar. 





BOOK IV. 


77 


For, because the triangles are similar, the angle ABC is 
equal to FGH ; and because the angle BCA is equal to GHF, 
and ACD to FHI, therefore the angle BCD is equal to GHI. 
For the same reason, the angle CDE is equal to HIK, and so 
on for the other angles. Therefore the two polygons are mu¬ 
tually equiangular. 

Moreover, the sides about the equal angles are proportion¬ 
al. For, because the triangles are similar, AB : FG : : BC : 
GH. Also, BC : GH : : AC : FH, and AC : FH : : CD : HI; 
hence BC : GH : : CD : HI. In the same manner, it may be 
proved that CD : HI : : DE : IK, and so on for the other 
sides. Therefore the two polygons are similar. 


PROPOSITION XXVI. THEOREM. 

The perimeters of similar polygons are to each other as their 
homologous sides; and their areas are as the squares of those 
sides. 



Let ABCDE, FGHIK 
be two similar polygons, 
and let AB be the side 33 
homologous to FG; then 
the perimeter of ABCDE 
is to the perimeter of 
FGHIK as AB is to FG ; 
and the area of ABCDE 
is to the area of FGHIK as AB a is to FG 3 . 

First. Because the polygon ABCDE is similar to the pol¬ 
ygon FGHIK (Def. 3), 

AB : FG : : BC : GH : : CD : HI, &c.; 
therefore (Prop. IX., B. II.) the sum of the antecedents AB 
-f BC-fCD, &c., which form the perimeter of the first figure, 
is to the sum of the consequents FG-f-GH+HI, &c., which 
form the perimeter of the second figure, as any one antece¬ 
dent is to its consequent, or as AB to FG. 

Secondly. Because the triangle ABC is similar to the tri¬ 
angle FGH, the triangle ABC : triangle FGH : : AC 3 : FH 3 
(Prop. XXIV.). 

And, because the triangle ACD is similar to the triangle 
FHI, 

ACD : FHI : : AC 3 : FH 3 . 

Therefore the triangle ABC : triangle FGH : : triangle 
ACD : triangle FHI (Prop. IV., B. II.). In the same man¬ 
ner, it may be proved that 

ACD : FHI: : ADE : FIK. 


78 


GEOMETRY. 


Therefore, as the sum of the antecedents ABC+ACD+ 
ADE, or the polygon ABODE, is to the sum of the conse¬ 
quents FGH+FHI+FIK, or the polygon FGHIK, so is any 
one antecedent, as ABC, to its consequent FGH; or, as AB* 
to FG*. Therefore, similar polygons, &c. 


PROPOSITION XXVII. THEOREM. 

If two chords in a circle intersect each other , the rectangle 
contained hy the parts of the one , is equal to the rectangle con¬ 
tained by the parts of the other. 

Let the two chords AB, CD in the circle 
ACBD, intersect each other in the point E; 
the rectangle contained by AE, EB is equal 
to the rectangle contained by DE, EC. 

Join AC and BD. Then, in the triangles 
ACE, DBE, the angles at E are equal, be¬ 
ing vertical angles (Prop. V., B. I.) ; the 
angle A is equal to the angle D, being in¬ 
scribed in the same segment (Prop. XV., Cor. 1., B. III.); 
therefore the angle C is equal to the angle B. The triangles 
are consequently similar; and hence (Prop. XVIII.) 

AE : DE : : EC : EB, 
or (Prop. I., B. II.), 

AE x EB =DE x EC. 

Therefore, if two chords, &c. 

Cor. The parts of two chords which intersect each other in 
a circle are reciprocally proportional; that is, AE : DE : ; 

EC : EB. 


PROPOSITION XXVIII. THEOREM. 

If from a point without a circle , a tangent and a secant be 
drawn , the square of the tangent will be equivalent to the rect¬ 
angle contained by the whole secant and its external segment. 

Let A be any point without the circle 
BCD, and let AB be a tangent, and AC a 
secant; then the square of AB is equiva¬ 
lent to the rectangle AD X AC. 

Join BD and BC. Then the triangles 
ABD and ABC are similar; because they 
have the angle A in common; also, the 
angle ABD formed by a tangent and a 
chord is measured by half the arc BD 


A 





BOOK IV. 


79 


(Prop. XVI., B. III.) ; and the angle C is measured by half 
the same arc, therefore the angle ABD is equal to C, and the 
two triangles ABD, ABC are equiangular, and, consequently, 
similar; therefore (Prop. XVIII.) 

AC : AB : : AB : AD; 
whence (Prop. I., B. II.), 

AB 2 =ACxAD. 

Therefore, if from a point, &c. 

Cor. 1 . If from a point without a circle, a tangent and a se¬ 
cant be drawn, the tangent will be a mean proportional be¬ 
tween the secant and its external segment. 

Cor. 2 . If from a point without a circle, two secants be 
drawn, the rectangles contained by the whole secants and 
their external segments will be equivalent to each other ; for 
each of these rectangles is equivalent to the square of the 
tangent from the same point. 

Cor. 3. If from a point without a circle, two secants be 
drawn, the whole secants will be reciprocally proportional to 
their external segments. 


PROPOSITION XXIX. THEOREM. 


If an angle of a triangle be bisected by a line which cuts the 
base , the rectangle contained by the sides of the triangle , is 
equivalent to the rectangle contained by the segments of the 
base , together with the square of the bisecting line. 


Let ABC be a triangle, and let the an¬ 
gle BAC be bisected by the straight line 
AD; the rectangle BAxAC is equiva¬ 
lent to BD X DC together with the square 
of AD. 

Describe the circle ACEB about the 
triangle, and produce AD to meet the cir¬ 
cumference in E, and join EC. Then, be¬ 
cause the angle BAD is equal to the an¬ 
gle CAE, and the angle ABD to the angle AEC, for they are 
in the same segment (Prop. XV., Cor. 1, B. III.), the trian¬ 
gles ABD, AEC are mutually equiangular and similar ; there¬ 
fore (Prop. XVIII.) 

BA : AD : : EA : AC; 



consequently (Prop. I., B. II.), 

BA X AC=AD X AE. 


But AE=AD-fDE ; and multiplying each of these equals 
by AD, we have (Prop. III.) ADx AE = AD 2 -f AD xDE. 
But ADxDE = BDxDC (Prop. XXVII.); hence 
B A x AC = BD x DC -1-AD 3 . 

Therefore, if an angle, &c. 



80 


GEOMETRY. 


PROPOSITION XXX. THEOREM. 


The rectangle contained by the diagonals of a quadrilateral 
inscribed in a circle, is equivalent to the sum of the rectangles 
of the opposite sides . 



Let ABCD be any quadrilateral in¬ 
scribed in a circle, and let the diagonals 
AC, BD be drawn; the rectangle AC x 
BD is equivalent to the sum of the two 
rectangles AD x BC and AB x CD. 

Draw the straight line BE, making the 
angle ABE equal to the angle DBC. To 
each of these equals add the angle EBD; 
then will the angle ABD be equal to the angle EBC. 
the angle BDA is equal to the angle BCE, because they are 
both in the same segment (Prop. XV., Cor. 1, B. III.) ; hence 
the triangle ABD is equiangular and similar to the triangle 
EBC. Therefore we have 

AD : BD : : CE : BC; 
and, consequently, ADxBC=BDxCE. 

Again, because the angle ABE is equal to the angle DBC, 
and the angle BAE to the angle BDC, being angles in the 
same segment, the triangle ABE is similar to the triangle 
DBC; and hence 


AB:AE::BD:CD; 
consequently, AB X CD =BD x AE. 

Adding together these two results, w^e obtain 

ADxBC+ABxCD=BDxCE+BDxAE, 
which equals BD x (Cfi+AE), or BD x AC. 
Therefore, the rectangle, &c. 


PROPOSITION XXXI. THEOREM. 

If from any angle of a triangle, a perpendicular be drawn 
to the opposite side or base, the rectangle contained by the sum 
and difference of the other two sides, is equivalent to the rect¬ 
angle contained by the sum and difference of the segments of 
the base. 

Let ABC be any triangle, and let AD be a perpendicular 
drawn from the angle A on the base BC; then 

(AC+AB) x (AC-AB) = (CD+DB) x (CD-DB). 

From A as a center, with a radius equal to AB, the short- 



BOOK IV. 


81 


C 

er of the two sides, describe a circumference BFE. Pro¬ 
duce AC to meet the circumference in E, and CB, if neces¬ 
sary, to meet it in F. 

Then, because AB is equal to AE or AG, CE=AC+AB, 
the sum of the sides ; and CG=AC—AB, the difference of the 
sides. Also, because BD is equal to DF (Prop. VI., B. III.); 
when the perpendicular falls within the triangle, CF=CD— 
DF=CD—DB, the difference of the segments of the base. 
But when the perpendicular falls without the triangle, CF= 
CD+DF=CD+DB, the sum of the segments of the base. 

Now in either case, the rectangle CExCG is equivalent 
to CBxCF (Prop. XXVIII., Cor. 2); that is, 

(AC + AB) x (AC - AB) = (CD + DB) x (CD - DB). 

Therefore, if from any angle, &c. 

Cor. If we reduce the preceding equation to a proportion 
(Prop. II., B. II.), we shall have 

BC : AC + AB : : AC-AB : CD-DB; 
that is, the longest side of the triangle, is to the sum of the 
two others, as the difference of the latter is to the difference 
of the segments made by the perpendicular. 




PROPOSITION XXXII. THEOREM. 


The diagonal and side of a square have no common measure . 

Let ABCD be a square, and AC its 
diagonal; AC and AB have no common 
measure. 

In order to find the common measure, 
if there is one, we must apply CB to CA 
as often as it is contained in it. For this 
purpose, from the center C, with a radius 
CB, describe the semicircle EBF. We 
perceive that CB is contained once in AC, with a remainder 
AE, which remainder must be compared with BC or its 
equal AB. 

Now, since the angle ABC is a right angle, AB is a tan¬ 
gent to the circumference; and AE : AB : : AB : AF (Prop. 

F 








82 


GEOMETRY. 


XXVIII., Cor. 1). Instead, therefore, of 
comparing AE with AB, we may substi¬ 
tute the equal ratio of AB to AF. But 
AB is contained twice in AF, with a re¬ 
mainder AE, which must be again com¬ 
pared with AB. Instead, however, of 
comparing AE with AB, we may again 
employ the equal ratio of AB to AF. A 
Hence at each operation we are obliged to compare AB with 
AF, which leaves a remainder AE ; from which we see that 
the process will never terminate, and therefore there is no 
common measure between the diagonal and side of a square; 
that is, there is no line which is contained an exact number 
of times in each of them. 






BOOK V. 


83 


BOOK V. 

PROBLEMS. 

Postulates. 

1 . A straight line may be drawn from any one point to 
any other point. 

2 . A terminated straight line may be produced to any 
length in a straight line. 

3. From the greater of two straight lines, a part may be 
cut off equal to the less. 

4. A circumference may be described from any center, and 
with any radius. 


PROBLEM i. 

To bisect a given straight line. 

Let AB be the given straight line 
which it is required to bisect. 

From the center A, with a radius great¬ 
er than the half of AB, describe an arc of A- . { . 
a circle (Postulate 4); and from the cen- \ j j 

ter B, with the same radius, describe an- 
other arc intersecting the former in D and x 

E. Through the points of intersection, draw the straight line 
DE (Post. 1); it will bisect AB in C. 

For, the two points D and E, being each equally distant 
from the extremities A and B, must both lie in the perpen¬ 
dicular, raised from the middle point of AB (Prop. XVIII., 
Cor., B. I.). Therefore the line DE divides the line AB 
into two equal parts at the point C. 



i 




84 


GEOMETRY. 


PROBLEM II. 

To draw a perpendicular to a straight line , from a given 
point in that line. 

Let BC be the given straight line, and A 
the point given in it; it is required to draw \ 
a straight line perpendicular to BC through ' 
the given point A. 

In the straight line BC take any point B, 

and make AC equal to AB (Post. 3). From ^ - Tq 

B as a center, with a radius greater than 
BA, describe an arc of a circle (Post. 4); and from C as a 
center, with the same radius, describe another arc intersect¬ 
ing the former in D. Draw AD (Post. 1), and it will be the 
perpendicular required. 

For, the points A and D, being equally distant from B and 
C, must be in a line perpendicular to the middle of BC (Prop. 
XVIII., Cor., B. I.). Therefore AD has been drawn per¬ 
pendicular to BC from the point A. 

Scholium. The same construction serves to make a right 
angle BAD at a given point A, on a given line BC. 


PROBLEM III. 


/ 


To draw a perpendicular to a straight line , from a given 
point without it. 

Let BD be a straight line of unlimited \ 

length, and let A be a given point without 
it. It is required to draw a perpendicular 
to BD from the point A. 

Take any point E upon the other side \. _ 

of BD; and from the center A, with the — 
radius AE, describe the arc BD cutting 
the line BCD in the two points B and D. 

From the points B and D as centers, de¬ 
scribe two arcs, as in Prob. II., cutting each other in F. 
Join AF, and it will be the perpendicular required. 

For the two points A and F are each equally distant from 
the points B and D; therefore the line AF has been drawn 
perpendicular to BD (Prop. XVIII., Cor., B. I.), from the 
given point A. 







BOOK V. 


85 


PROBLEM IV. 



At a given point in a straight line , to make an angle equal 
to a given angle. 

Let AB be the given straight 
line, A the given point in it, and 
’ C the given angle ; it is required 
to make an angle at the point A 
in the straight line AB, that shall 
be equal to the given angle C. 

With C as a center, and any radius, describe an arc DE 
terminating in the sides of the angle; and from the point A 
as a center, with the same radius, describe the indefinite arc 
BF. Draw the chord DE; and from B as a center, with a 
radius equal to DE, describe an arc cutting the arc BF in G. 
Draw AG, and the angle BAG will be equal to the given 
angle C. 

For the two arcs BG, DE are described with equal radii, 
and they have equal chords ; they are, therefore, equal (Prop. 

III. , B. III.). But equal arcs subtend equal angles (Prop. 

IV. , B. III.); and hence the angle A has been made equal to 
the given angle C. 


problem v. 

To bisect a given arc or angle. 

First. Let ADB be the given arc which c 

it is required to bisect. 

Draw the chord AB, and from the center 
C draw CD perpendicular to AB (Prob. 

III.); it will bisect the arc ADB (Prop. 

VI., B. III.), because CD is a radius per¬ 
pendicular to a chord. 

Secondly. Let ACB be an angle which it is required to bi¬ 
sect. From C as a center, with any radius, describe an arc 
AB; and, by the first case, draw the line CD bisecting the 
arc ADB. The line CD will also bisect the angle ACB. For 
the angles ACD, BCD are equal, being subtended by the 
equal arcs AD, DB (Prop. IV., B. III.). 

Scholium. By the same construction, each of the halves 
AD, DB may be bisected; and thus by successive bisections, 
an arc or angle may be divided into four equal parts, into 
eight, sixteen, &c. 







86 


GEOMETRY. 


PROBLEM VI. 

Through a given point , to draw a straight line parallel to 
a given line. 

Let A be the given point, and BC the B DC 

given straight line; it is required to draw f? . 

through the point A, a straight line paral- 

lel t0 BC * - a~ -E 

In BC take any point D, and join AD. 

Then at the point A, in the straight line AD, make the angle 
DAE equal to the angle ADB (Prob. IV.). 

Now, because the straight line AD, which meets the two 
straight lines BC, AE, makes the alternate angles ADB, DAE 
equal to each other, AE is parallel to BC (Prop. XXII., B. 
I.). Therefore the straight line AE has been drawn through 
the point A, parallel to the given line BC. 


PROBLEM VII. 

Two angles of a triangle being given , to find the third angle. 

The three angles of every triangle are to- jy 
gether equal to two right angles (Prop. 

XXVII., B. I.). Therefore, draw the in- 

definite line ABC. At the point B make _ 

the angle ABD equal to one of the given ABC 
angles (Prob. IV.), and the angle DBE equal to the other 
given angle; then will the angle EBC be equal to the third 
angle of the triangle. For the three angles ADB, DBE, 
EBC are together equal to two right angles (Prop. II., B. 
I.), which is the sum of all the angles of the triangle. 



PROBLEM VIII. 

Given two sides and the included angle of a triangle, to con¬ 
struct the triangle. 

Draw the straight line BC equal to one A 

of the given sides. At the point B make 
the angle ABC equal to the given angle 
(Prob. IV.); and take AB equal to the other 
given side. Join AC, and ABC will be the b 


C 







BOOK V. 


87 


triangle required. For its sides AB, BC are made equal to 
the given sides, and the included angle B is made equal to 
the given angle. 


PROBLEM IX. 

Given one side and two angles of a triangle , to construct the 
triangle. 

The two given angles will either be both adjacent to the 
given side, or one adjacent and the other opposite. In the 
latter case, find the third angle (Prob. VII.) ; and then the 
two adjacent angles will be known. 

Draw the straight line AB equal to the 
given side; at the point A make the angle 
BAC equal to one of the adjacent angles; 
and at the point B make the angle ABD 
equal to the other adjacent angle. The two 
lines AC, BD will cut each other in E, and A B 

ABE will be the triangle required ; for its side AB is equal 
to the given side, and two of its angles are equal to the given 
angles. 



problem x. 

Given the three sides of a triangle , to construct the triangle. 

Draw the straight line BC equal to one of 
the given sides. From the point B as a cen¬ 
ter, with a radius equal to one of the other 
sides, describe an arc of a circle; and from 
the point C as a center, with a radius equal 
to the third side, describe another arc cutting 
the former in A. Draw AB, AC; then will 
ABC be the triangle required, because its three sides are 
equal to the three given straight lines. 

Scholium. If one of the given lines was greater than the 
sum of the other two, the arcs would not intersect each other, 
and the problem would be impossible; but the solution will 
always be possible, when the sum of any two sides is greater 
than the third. 





88 


GEOMETRY. 


PROBLEM XI. 



Given two sides of a triangle , and an angle opposite one of 
them , to construct the triangle. 

Draw an indefinite straight line 
BC. At the point B make the angle 
ABC equal to the given angle, and 
make BA equal to that side which is 
adjacent to the given angle. Then . 

from A as a center, with a radius E "-- - 

equal to the other side, describe an arc cutting BC in the 
points E and F. Join AE, AF. If the points E and F both 
fall on the same side of the angle B, each of the triangles 
ABE, ABF will satisfy the given conditions ; but if they fall 
upon different sides of B, only one of them, as ABF, will 
satisfy the conditions, and therefore this will be the triangle 
required. 

If the points E and F coincide with one another, which 
will happen when AEB is a right angle, there will be only 
one triangle ABD, which is the triangle required. 

Scholium. If the side opposite the given angle were less 
than the perpendicular let fall from A upon BC, the problem 
would be impossible. 


PROBLEM XII. 


Given two adjacent sides of a parallelogram , and the in¬ 
cluded angle , to construct the parallelogram. 

Draw the straight line AB equal to 

one of the given sides. At the point A ^_ 

make the angle BAC equal to the / 

given angle; and take AC equal to / 

the other given side. From the point - 

C as a center, with a radius equal to A B 

AB, describe an arc; and from the point B as a center, with 
a radius equal to AC, describe another arc intersecting the 
former in D. Draw BD, CD; then will ABDC be the paral¬ 
lelogram required. 

For, by construction, the opposite sides are equal; there¬ 
fore the figure is a parallelogram (Prop. XXX., B. I.), and it. 
is formed with the given sides and the given angle. 






BOOK V. 


89 


Cor. If the given angle is a right angle, the figure will be 
a rectangle; and if, at the same time, the sides are equal, it 
will be a square. 


PROBLEM XIII. 

To find the center of a given circle or arc. 

Let ABC be the given circle or arc; ^ - 

it is required to find its center. / 

Take any three points in the arc, as / \ 

A, B, C, and join AB, BC. Bisect AB ( £ ] 

in D (Prob. I.), and through D draw DF 
perpendicular to AB (Prob. II.). In the W )k/ 
same manner, draw EF perpendicular to 
BC at its middle point. The perpen- B 

diculars DF, EF will meet in a point F equally distant from 
the points A, B, and C (Prop. VII., B. III.) ; and therefore F is 
the center of the circle. 

Scholium. By the same construction, a circumference may 
be made to pass through three given points A, B, C; and 
also, a circle may be described about a triangle. 


PROBLEM XIV. 



Through a given point , to draw a tangent to a given circle. 

First. Let the given point A be 
without the circle BDE ; it is re¬ 
quired to draw a tangent to the cir¬ 
cle through the point A. 

Find the center of the circle C, and 
join AC. Bisect AC in D ; and with 
D as a center, and a radius equal to 
AD, describe a circumference intersecting the given circum¬ 
ference in B. Draw AB, and it will be the tangent required. 

Draw the radius CB. The angle ABC, being inscribed in 
a semicircle, is a right angle (Prop. XV., Cor. 2, B. III.). 
Hence the line AB is a perpendicular at the extremity of the 
radius CB; it is, therefore, a tangent to the circumference 
(Prop. IX., B. III.). 

Secondly. If the given point is in the circumference of the 
circle, as the point B, draw the radius BC, and make BA 
perpendicular to BC. BA will be the tangent required 
(Prop. IX., B. III.). 






90 


GEOMETRY. 


Scholium. When the point A lies without the circle, two 
tangents may always be drawn; for the circumference whose 
center is D intersects the given circumference in two points. 


problem xv. 

To inscribe a circle in a given triangle . 

Let ABC be the given triangle; it is 
required to inscribe a circle in it. 

Bisect the angles B and C by the 
lines BD, CD, meeting each other in 
the point D. From the point of inter¬ 
section, let fall the perpendiculars DE, 

DF, DG on the three sides of the tri¬ 
angle ; these perpendiculars will all be 
equal. For, by construction, the angle ® 

EBD is equal to the angle FBD; the right angle DEB is 
equal to the right angle DFB; hence the third angle BDE 
is equal to the third angle BDF (Prop. XXVII., Cor. 2, B. 
I.). Moreover, the side BD is common to the two triangles 
BDE, BDF, and the angles adjacent to the equal side are 
equal; therefore the two triangles are equal, and DE is equal 
to DF. For the same reason, DG is equal to DF. There¬ 
fore the three straight lines DE, DF, DG are equal to each 
other; and if a circumference be described from the center 
D, with a radius equal to DE, it will pass through the ex¬ 
tremities of the lines DF, DG. It will also touch the straight 
lines AB, BC, CA, because the angles at the points E, F, G 
are right angles (Prop. IX., B. III.). Therefore the circle 
EFG is inscribed in the triangle ABC (Def. 11, B. III.) 

Scholium. The three lines which bisect the angles of a 
triangle, all meet in the same point, viz., the center of the in¬ 
scribed circle. 



problem xvi. 

Upon a given straight line, to describe a segment of a circle 
which shall contain a given angle. 

Let AB be the given straight line, upon which it is re¬ 
quired to describe a segment of a circle containing a given 
angle. 

At the point A, in the straight line AB, make the angle 
BAD equal to the given angle; and from the point A draw 



BOOK V. 


91 



draw EC perpendicular to AB. From the point C, where 
these perpendiculars meet, with a radius equal to AC, de¬ 
scribe a circle. Then will AGB be the segment required. 

For, since AD is a perpendicular at the extremity of the 
radius AC, it is a tangent (Prop. IX., B. III.); and the angle 
BAD is measured by half the arc AFB (Prop. XVI., B. III.). 
Also, the angle AGB, being an inscribed angle, is measured 
by half the same arc AFB; hence the angle AGB is equal to 
the angle BAD, which, by construction, is equal to the given 
angle. Therefore all the angles inscribed in the segment 
AGB are equal to the given angle. 

Scholium. If the given angle was a right angle, the re¬ 
quired segment would be a semicircle, described on AB as a 
diameter. 


PROBLEM XVII. 

To divide a given straight line into any number of equal 
parts , or into parts proportional to given lines. 

First. Let AB be the given straight 
line which it is proposed to divide into 
any number of equal parts, as, for ex¬ 
ample, five. 

From the point A draw the indefinite 
straight line AC, making any angle 
with AB. In AC take any point D, A E E 

and set off AD five times upon AC. Join BC, and draw DE 
parallel to it; then is AE the fifth part of AB. 

For, since ED is parallel to BC, AE : AB : : AD : AC 
(Prop. XVI., B. IV.). But AD is the fifth part of AC; 
therefore AE is the fifth part of AB. 

Secondly. Let AB -be the given straight line, and AC a di¬ 
vided line; it is required to divide AB similarly to AC. Sup¬ 
pose AC to be divided in the points D and E. Place AB, 
AC so as to contain any angle ; join BC, and through the 








92 


GEOMETRY. 


points D, E draw DF, EG parallel to BC. 
The*line AB will be divided into parts 
proportional to those of AC. 

For, because DF and EG are both par¬ 
allel to CB, we have AD : AF :: DE : FG 
:: EC: GB (Prop. XVI., Cor. 2, B. IV.). 



PROBLEM XVIII. 


To find a fourth proportional to three given lines. 

From any point A draw two straight 
lines AD, AE, containing any angle 
DAE; and make AB, BD, AC respect¬ 
ively equal to the proposed lines. Join 
B, C; and through D draw DE parallel 
to BC; then will CE be the fourth pro¬ 
portional required. 

For, because BC is parallel to DE, we have 

AB : BD : : AC : CE (Prop. XVI., B. IV.). 

Cor . In the same manner may be found a third propor¬ 
tional to two given lines A and B; for this will be the same 
as a fourth proportional to the three lines A, B. B. 



PROBLEM XIX. 


To find a mean proportional between two given lines. 


_P 


\ 


Let AB, BC be the two given straight 
lines; it is required to find a mean pro¬ 
portional between them. , 

Place AB, BC in a straight line; upon / 

AC describe the semicircle ADC; and L 
from the point B draw BD perpendicular ^ B G 

to AC. Then will BD be the mean proportional required. 

For the perpendicular BD, let fall from a point in the cir¬ 
cumference upon the diameter, is a mean proportional be¬ 
tween the two segments of the diameter AB, BC (Prop. 
XXII., Cor., B. IV.); and these segments are equal to the 
two given lines. 






BOOK V. 


93 


PROBLEM XX. 

To divide a given line into two parts , such that the greater 
part may be a mean proportional between the whole line and 
the other part. 

Let AB be the given straight line; 
it is required to divide it into two 
parts at the point F, such that AB : 

AF : : AF : FB. 

At the extremity of the line AB, N / 

erect the perpendicular BC, and make - i —'' 

it equal to the half of AB. From C A F B 
as a center, with a radius equal to CB, describe a circle. 
Draw AC cutting the circumference in D; and make AF 
equal to AD. The line AB will be divided in the point F in 
the manner required. 

For, since AB is a perpendicular to the radius CB at its ex¬ 
tremity, it is a tangent (Prop. IX., B. III.) ; and if we pro¬ 
duce AC to E, we shall have AE ; AB : : AB : AD (Prop. 
XXVIII., B. IV.). Therefore, by division (Prop. VII., B. 
II.), AE —AB : AB : : AB—AD : AD. But, by construction, 
AB is equal to DE; and therefore AE—AB is equal to AD 
or AF; and AB—AD is equal to FB. Hence AF : AB : : 
FB : AD or AF; and, consequently, by inversion (Prop. V., 
B. II.), 

AB : AF : : AF : FB. 

Scholium. The line AB is said to be divided in extreme 
and mean ratio. An example of its use may be seen in Prop. 
V., Book VI. 



PROBLEM XXI. 


Through a given point in a given angle , to draw a straight 
line so that the parts included between the point and the sides 
of the angle, may be equal. 


Let A be the given point, and BCD the 
given angle; it is required to draw through 
A a line BD, so that BA may be equal to 
AD. 

Through the point A draw AE parallel to 
BC ; and take DE equal to CE. Through 
the points D and A draw the line BAD; it 
will be the line required. 







94 


GEOMETRY. 


For, because AE is parallel to BC, we have (Prop. XVI., 
B. IV.), 

DE : EC : : DA : AB. 

But DE is equal to EC ; therefore DA is equal to AB. 


PROBLEM XXII. 


To describe a square that shall be equivalent to a given 
parallelogram , or to a given triangle. 

First. Let ABDC be the given paral¬ 
lelogram, AB its base, and CE its altitude. 

Find a mean proportional between AB and 
CE (Prob. XIX.), and represent it by X; 


D 


the square described on X will be equiva- A E B 

lent to the given parallelogram ABDC. 

For, by construction, AB : X :: X : CE ; hence X 2 is equal 
to AB X CE (Prop. I., Cor., B. II.). But AB X CE is the 
measure of the parallelogram; and X 2 is the measure of the 
square. Therefore the square described on X is equivalent 
to the given parallelogram ABDC. 

Secondly. Let ABC be the given triangle, A 

BC its base, and AD its altitude. Find a 
mean proportional between BC and the half 
of AD, and represent it by X. Then will 
the square described on X be equivalent to 
the triangle ABC. B DC 

For, by construction, BC : X : : X : \ AD ; hence X 2 is 
equivalent to BC X | AD. But BC X | AD is the measure of 
the triangle ABC; therefore the square described on X is 
equivalent to the triangle ABC. 



PROBLEM XXIII. 

Upon a given line , to construct a rectangle equivalent to a 
given rectangle. 

Let AB be the given straight 
line, and CDFE the given rect¬ 
angle. It is required to con¬ 
struct on the line AB a rectan¬ 
gle equivalent to CDFE. 

Find a fourth proportional 
(Prob. XVIII.) to the three lines AB, CD, CE, and let AG 
be that fourth proportional. The rectangle constructed on 
the lines AB, AG will be equivalent to CDFE. 











BOOK V. 


95 


For, because AB : CD : : CE : AG, by Prop. I., B. II., 
ABxAG=CDxCE. Therefore the rectangle ABHG is 
equivalent to the rectangle CDFE ; and it is constructed 
upon the given line AB. 


PROBLEM XXIV. 

To construct a triangle which shall he equivalent to a given 
polygon. 

Let ABODE be the given polygon; it 
is required to construct a triangle equiva¬ 
lent to it. 

Draw the diagonal BD cutting off the 
triangle BCD. Through the point C, 
draw CF parallel to DB, meeting AB 
produced in F. Join DF; and the poly¬ 
gon AFDE will be equivalent to the polygon ABCDE. 

For the triangles BFD, BCD, being upon the same base 
BD, and between the same parallels BD, FC, are equivalent. 
To each of these equals, add the polygon ABDE; then will 
the polygon AFDE be equivalent to the polygon ABCDE; 
that is, we have found a polygon equivalent to the given 
polygon, and having the number of its sides diminished by 
one. 

In the same manner, a polygon may be found equivalent 
to AFDE, and having the number of its sides diminished by 
one; and, by continuing the process, the number of sides 
may be at last reduced to three, and a triangle be thus obtain¬ 
ed equivalent to the given polygon. 



PROBLEM XXV. 


To make a square equivalent to the sum or difference of two 
given squares. 

First. To make a square equivalent to the sum of two 
given squares. Draw two indefinite lines c 

AB, BC at right angles to each other. Take 
AB equal to the side of one of the given 
squares, and BC equal to the side of the 
other. Join AC; it will be the side of the A 
required square. 

For the triangle ABC, being right-angled at B, the square 






96 


GEOMETRY. 


on AC will be equivalent to the sum of the squares upon AB 
and BC (Prop. XI., B. IV.). 

Secondly. To make a square equivalent to the difference 
of two given squares. Draw the lines AB, BC at right an¬ 
gles to each other; and take AB equal to the side of the less 
square. Then from A as a center, with a radius equal to the 
side of the other square, describe an arc intersecting BC in 
C; BC will be the side of the square required; because the 
square of BC is equivalent to the difference of the squares of 
AC and AB (Prop. XI., Cor. 1, B. IV.). 

Scholium. In the same manner, a square may be made 
equivalent to the sum of three or more given squares; for 
the same construction which reduces two of them to one, 
will reduce three of them to two, and these two to one. 


PROBLEM XXVI. 

Upon a given straight line , to construct a polygon similar 
to a given polygon. 

Let ABCDE be the giv¬ 
en polygon, and FG be 
the given straight line; it E 
is required upon the line 
FG to construct a poly¬ 
gon similar to ABCDE. 

Draw the diagonals BD, 

BE. At the point F, in 
the straight line FG, make the angle GFK equal to the angle 
BAE; and at the point G make the angle FGK equal to the 
angle ABE. The lines FK, GK will intersect in K, and 
FGK will be a triangle similar to ABE. In the same man¬ 
lier, on GK construct the triangle GKI similar to BED, and 
on GI construct the triangle GIH similar to BDC. The 
polygon FGHIK will be the polygon required. For these 
two polygons are composed of the same number of triangles, 
which are similar to each other, and similarly situated ; there¬ 
fore the polygons are similar (Prop. XXV., Cor., B. IV.) 



PROBLEM XXVII. 

Given the area of a rectangle , and the sum of its adjacent 
sides , to construct the rectangle. 

Let AB be a straight line equal to the sum of the sides of 
the required rectangle. 





BOOK V. 


97 


Upon AB as a diameter, describe a c 
semicircle. At the point A erect the 
perpendicular AC, and make it equal to 
the side of a square having the given 
area. Through C draw the line CD par- A E B 

allel to AB, and let it meet the circumference in D; and 
from D draw DE perpendicular to AB. Then will AE and 
EB be the sides of the rectangle required. 

For, by Prop. XXII., Cor., B. IV., the rectangle AExEB 
is equivalent to the square of DE or CA, which is, by con¬ 
struction, equivalent to the given area. Also, the sum of the 
sides AE and EB is equal to the given line AB. 

Scholium . The side of the square having the given area, 
must not be greater than the half of AB ; for in that case the 
line CD would not meet the circumference ADB. 



PROBLEM XXVIII. 

Given the area of a rectangle , and the difference of its ad¬ 
jacent sides , to construct the rectangle . 

Let AB be a straight line equal to the c 
difference of the sides of the required rect¬ 
angle. 

Upon AB as a diameter, describe a cir¬ 
cle ; and at the extremity of the diameter, 
draw the tangent AC equal to the side of 
a square having the given area. Through 
the point C and the center F draw the 
secant CE; then will CD, CE be the adjacent sides of the 
rectangle required. 

For, by Prop. XXVIII., B. IV., the rectangle CDxCE is 
equivalent to the square of AC, which is, by construction, 
equivalent to the given area. Also, the difference of the lines 
CE, CD is equal to DE or AB. 

G 








/ 











98 


GEOMETRY. 


BOOK VI. 

REGULAR POLYGONS, AND THE AREA OF THE CIRCLE. 
Definition. 

A regular polygon is one which is both equiangular and 
equilateral. 

An equilateral triangle is a regular polygon of three sides ; 
a square is one of four. 


PROPOSITION I. THEOREM. 

Regular polygons of the same number of sides are similar 
figures. 

Let ABCDEF, abedef be 
two regular polygons of the 
same number of sides ; then 
will they be similar figures. 

For, since the two polygons 
have the same number of 
sides, they must have the 
same number of angles. Moreover, the sum of the angles of 
the one polygon is equal to the sum of the angles of the other 
(Prop. XXVIII., B. I.); and since the polygons are each 
equiangular, it follows that the angle A is the same part of 
the sum of the angles A, B, C, D, E, F, that the angle a is 
of the sum of the angles a, b, c, d, e, f. Therefore the two 
angles A and a are equal to each other. The same is true 
of the angles B and b , C and c, &c. 

Moreover, since the polygons are regular, the sides AB, 
BC, CD, &c., are equal to each other (Def.) ; so, also, are the 
sides ab y be , cd y &c. Therefore AB : ab :: BC : be :: CD : cd , 
&c. Hence the two polygons have their angles equal, and 
their homologous sides proportional; they are consequently 
similar (Def. 3, B. IV.). Therefore, regular polygons, &c. 

Cor. The perimeters of two regular polygons of the same 
number of sides, are to each other as their homologous sides, 






BOOK VI. 


99 


and their areas are as the squares of those sides (Prop. 
XXVI., B. IV.). 

Scholium. The angles of a regular polygon are deter¬ 
mined by the number of its sides. 


PROPOSITION II. THEOREM. 

A circle may he described about any regular polygon , and 
another may be inscribed within it. 

Let ABCDEF be any regular polygon; 
a circle may be described about it, and 
another may be inscribed within it. 

Bisect the angles FAB, ABC by the 
straight lines AO, BO ; and from the point 
O in which they meet, draw the lines OC, 

OD, OE, OF to the other angles of the 
polygon. 

Then, because in the triangles OBA, OBC, AB is, by 
hypothesis, equal to BC, BO is common to the two triangles, 
and the included angles OBA, OBC are, by construction, 
equal to each other; therefore the angle OAB is equal to the 
angle OCB. But OAB is, by construction, the half of FAB ; 
and FAB is, by hypothesis, equal to DCB; therefore OCB is 
the half of DCB; that is, the angle BCD is bisected by the 
line OC. In the same manner it may be proved that the an¬ 
gles CDE, DEF, EFA are bisected by the straight lines OD, 

OE, OF. 

Now because the angles OAB, OBA, being halves of equal 
angles, are equal to each other, OA is equal to OB (Prop. 
XI., B. I.). For the same reason, OC, OD, OE, OF are each 
of them equal to OA. Therefore a circumference described 
frqm the center O, with a radius equal to OA, will pass 
through each of the points B, C, D, E, F, and be described 
about the polygon. 

Secondly. A circle may be inscribed within the polygon 
ABCDEF. For the sides AB, BC, CD, &c., are equal 
chords of the same circle; hence they are equally distant 
from the center O (Prop. VIII., B. III.); that is, the perpen¬ 
diculars OG, OH, &c., are all equal to each other. There¬ 
fore, if from O as a center, with a radius OG, a circumference 
be described, it will touch the side BC (Prop. IX., B. III.), 
and each of the other sides of the polygon ; hence the circle 
will be inscribed within the polygon. Therefore a circle 
may be described, &c. 

Scholium 1 . In regular polygons, the center of the inscribed 






100 


GEOMETRY. 


and circumscribed circles, is also called the center of the poly¬ 
gon; and the‘perpendicular from the center upon one of the 
sides, that is, the radius of the inscribed circle, is called the 
apothem of the polygon. 

Since all the chords AB, BC, &c., are equal, the angles at 
the center, AOB, BOC, &c., are equal; and the value of each 
may be found by dividing four right angles by the number 
of sides of the polygon. 

Scholium 2. To inscribe a regular polygon of any number 
of sides in a circle, it is only necessary to divide the circum* 
ference into the same number of equal parts ; for, if the 
arcs are equal, the chords AB, BC, CD, &c., will be equal. 
Hence the triangles AOB, BOC, COD, &c., will also be 
equal, because they are mutually equilateral; therefore all 
the angles ABC, BCD, CDE, &c., will be equal, and the 
figure ABCDEF will be a regular polygon. 


PROPOSITION III. PROBLEM. 

To inscribe a square in a given circle. 

Let ABCD be the given circle ; it is re¬ 
quired to inscribe a square in it. 

Draw two diameters AC, BD at right 
angles to each other; and join AB, BC, 

CD, DA. Because the angles AEB, BEC, 

&c., are equal, the chords AB, BC, &c., 
are also equal. And because the angles 
ABC, BCD, &c., are inscribed in semicir¬ 
cles, they are right angles (Prop. XV., Cor. 2, B. III.). 
Therefore ABCD is a square, and it is inscribed in the circle 
ABCD. 

Cor . Since the triangle AEB is right-angled and isosceles, 
we have the proportion, AB : AE : : ^/2 : 1 (Prop. XI., Cor. 
3, B. IV.) ; therefore the side of the inscribed square is to the 
radius , as the square root of 2 is to unity. 



PROPOSITION IV. THEOREM. 

t The side of a regular hexagon is equal to the radius of the 
circumscribed circle. 

Let ABCDEF be a regular hexagon inscribed in a circle 
whose center is O; then any side as AB will be equal to the 
radius AO. 





BOOK VI. 


101 


Draw the radius BO. Then the angle 
AOB is the sixth part of four right angles 
(Prop. II.. Sch. 1), or the third part of two 
right angles. Also, because the three an¬ 
gles of every triangle are equal to two 
right angles, the two angles OAB, OBA 
are together equal to two thirds of two 
right angles; and since AO is equal to BO, each of these an¬ 
gles is one third of two right angles. Hence the triangle 
AOB is equiangular, and AB is equal to AO. Therefore the 
side of a regular hexagon, &c. 

Cor. To inscribe a regular hexagon in a given circle, tlie 
radius must be applied six times upon the circumference. 
By joining the alternate angles A, C, E, an equilateral trian¬ 
gle will be inscribed in the circle. 



PROPOSITION V. PROBLEM. 

To inscribe a regular decagon in a given circle . 

Let ABF be the given circle; it is re- p 

quired to inscribe in it a regular decagon. 

Take C the center of the circle; draw 
the radius AC, and divide it in extreme 
and mean ratio (Prob. XX., B. V.) at 
the point D. Make the chord AB equal 
to CD the greater segment; then will 
AB be the side of a regular decagon in¬ 
scribed in the circle. 

Join BC, BD. Then, by construction, 

AC : CD : : CD : AD; but AB is equal to CD; therefore 
AC : AB : : AB : AD. Hence the triangles ACB, ABD 
have a common angle A included between proportional 
sides; they are therefore similar (Prop. XX., B. IV.) And 
because the triangle ACB is isosceles, the triangle ABD must 
also be isosceles, and AB is equal to BD. But AB was made 
equal to CD; hence BD is equal to CD, and the angle DBC 
is equal to the angle DCB. Therefore the exterior angle 
ADB, which is equal to the sum of DCB and DBC, must be 
double of DCB. But the angle ADB is equal to DAB; there¬ 
fore each of the angles CAB, CBA is double of the angle 
ACB. Hence the sum of the three angles of the triangle 
ACB is five times the angle C. But these three angles are 
equal to two right angles (Prop. XXVII., B. I.); therefore 
the angle C is the fifth part of two right angles, or the tenth 
part of four right angles. Hence the arc AB is one tenth of 







102 


GEOMETRY. 


the circumference, and the chord AB is the side of a regular 
decagon inscribed in the circle. 

Cor. 1 . By joining the alternate angles of the regular dec¬ 
agon, a regular pentagon may be inscribed in the circle. 

Cor. 2. By combining this Proposition with the preceding, 
a regular pentedecagon may be inscribed in a circle. 

For, let AE be the side of a regular hexagon ; then the arc 
AE will be one sixth of the whole circumference, and the arc 
AB one tenth of the whole circumference. Hence the arc 
BE will be ± — T V or and the chord of this arc will be the 
side of a regular pentedecagon. 

Scholium. By bisecting the arcs subtended by the sides of 
any polygon, another polygon of double the number of sides 
may be inscribed in a circle. Hence the square will enable 
us to inscribe regular polygons of 8, 16, 32, &c., sides; the 
hexagon will enable us to inscribe polygons of 12, 24, &c., 
sides ; the decagon will enable us to inscribe polygons of 
20, 40, &c., sides ; and the pentedecagon, polygons of 30, 60, 
&c., sides. 

The ancient geometricians were unacquainted with any 
method of inscribing in a circle, regular polygons of 7, 9, 11, 
13, 14, 17, &c., sides; and for a long time it was believed 
that these polygons could not be constructed geometrically ; 
but Gauss, a German mathematician, has shown that a regu¬ 
lar polygon of 17 sides may be inscribed in a circle, by em¬ 
ploying straight lines and circles only. 

PROPOSITION VI. PROBLEM. 

A regular polygon inscribed • in a circle being given , to de¬ 
scribe a similar polygon about the circle. 

Let ABCDEF be a regular polygon 
inscribed in the circle ABD ; it is re¬ 
quired to describe a similar polygon 
about the circle. 

Bisect the arc AB in G, and through L 
G draw the tangent LM. Bisect also 
the arc BC in H, and through H draw 
the tangent MN, and in the same man¬ 
ner draw tangents to the middle points 
of the arcs Cl), DE, &c. These tangents, by their intersec¬ 
tions, will form a circumscribed polygon similar to the one 
inscribed. 

Find O the center of the circle, and draw the radii OG, 
OH. Then, because OG is perpendicular to the tangent LM 
(Prop. IX., B. III.), and also to the chord AB (Prop. VI., 







BOOK VI. 


103 

Sch., B. III.), the tangent is parallel to the chord (Prop. XX., 
B. I.). In the same manner it may be proved that the other 
sides of the circumscribed polygon are parallel to the sides 
of the inscribed polygon; and therefore the angles of the 
circumscribed polygon are equal to those of the inscribed one 
(Prop. XXVI., B. I.). 

Since the arcs BG, BH are halves of the equal arcs AGB, 
BHC, they are equal to each other; that is, the vertex B is 
at the middle point of the arc GBH. Join OM ; the line OM 
will pass through the point B. For the right-angled trian¬ 
gles OMH, OMG have the hypothenuse OM common, and 
the side OH equal to OG; therefore the angle GOM is equal 
to the angle HOM (Prop. XIX., B. I.), and the line OM passes 
through the point B, the middle of the arc GBH. 

Now because the triangle OAB is similar to the triangle 
OLM, and the triangle OBC to the triangle OMN, we have 
the proportions 

AB : LM : : BO : MO; 
also, BC : MN : : BO : MO ; 

therefore (Prop. IV., B. II.), 

AB : LM : : BC : MN. 

But AB is equal to BC ; therefore LM is equal to MN. In 
the same manner, it may be proved that the other sides of the 
circumscribed polygon are equal to each other. Hence this 
polygon is regular, and similar to the one inscribed. 

Cor. 1 . Conversely, if the circumscribed polygon is given, 
and it is required to form the similar inscribed one, draw the 
lines OL, OM, ON, &c., to the angles of the polygon ; these 
lines will meet the circumference in the points A, B, C, &c. 
Join these points by the lines AB, BC, CD, &c., and a similar 
polygon will be inscribed in the circle. 

Or we may simply join the points of contact G, H, I, &c., 
by the chords GH, HI, &c., and there will be formed an in¬ 
scribed polygon similar to the circumscribed one. 

Cor. 2. Hence we can circumscribe about a circle, any 
regular polygon which can be inscribed within it, and con¬ 
versely. 

Cor. 3. A side of the circumscribed polygon MN is equal 
to twice MH, or MG + MH. 


PROPOSITION VII. THEOREM. 

The area of a regular polygon is equivalent to the product 
of its perimeter , by half the radius of the inscribed circle. 

Let ABCDEF be a regular polygon, and G the center of 


104 


GEOMETRY. 


the inscribed circle. From G draw 
lines to all the angles of the polygon. 

The polygon will thus be divided into 
as many triangles as it has sides; and 
the common altitude of these triangles 
is GH, the radius of the circle. Now, 
the area of the triangle BGC is equal to 
the product of BC by the half of GH 
(Prop. VI., B. IV.); and so of all the 
other triangles having their vertices in G. Hence the sum 
of all the triangles, that is, the surface of the polygon, is 
equivalent to the product of the sum of the bases AB, BC, 
&c.; that is, the perimeter of the polygon, multiplied by half 
of GH, or half the radius of the inscribed circle. Therefore, 
the area of a regular polygon, &c. 



PROPOSITION VIII. THEOREM. 


The perimeters of two regular polygons of the same number 
of sides, are as the radii of the inscribed or circumscribed cir¬ 
cles, and their surfaces are as the squares of the radii. 



Let ABCDEF, abcdef be 
two regular polygons of the 
same number of sides ; let G 
and g be the centers of the 
circumscribed circles; and 
let GH, gh be drawn per¬ 
pendicular to BC and be ; -^C 

then will the perimeters of the polygons be as the radii BG, 
bg; and, also, as GH, gh, the radii of the inscribed circles. 

The angle BGC is equal to the angle bgc (Prop. II., Sch. 
1); and since the triangles BGC, bgc are isosceles, they are 
similar. So, also, are the right-angled triangles BGH,*6gA; 
and, consequently, BC : be : : BG : bg : : GH : gh. But the 
perimeters of the two polygons are to each other as the sides 
BC, be (Prop. I., Cor.) ; they are, therefore, to each other as 
the radii BG, bg of the circumscribed circles; and also as the 
radii GH, gh of the inscribed circles. 

The surfaces of these polygons are to each other as the 
squares of the homologous sides BC, be (Prop. I., Cor.); they 
are, therefore, as the squares of BG, bg, the radii of the cir¬ 
cumscribed circles ; or as the squares of GH, gh, the radii of 
the inscribed circles. 











BOOK VI. 


105 


PROPOSITION IX. PROBLEM. 



The surface of a regular inscribed polygon , and that of a 
similar circumscribed polygon, being given; to find the surfaces 
of regular inscribed and circumscribed polygons having double 
the number of sides. 

Let AB be a side of the given in 
scribed polygon; EF parallel to AB, a 
side of the similar circumscribed poly¬ 
gon; and C the center of the circle. 

Draw the chord AG, and it will be the 
side of the inscribed polygon having 
double the member of sides. At the 
points A and B draw tangents, meeting 
EF in the points H and I; then will 
HI, which is double of HG, be a side of 
the similar circumscribed polygon (Prop. VI., Cor. 1). Let 
p represent the inscribed- polygon whose side is AB, P the 
corresponding circumscribed polygon; p r the inscribed poly¬ 
gon having double the number of sides, P' the similar cir¬ 
cumscribed polygon. Then it is plain that the space CAD is 
the same part of p , that CEG is of P ; also, CAG of p r , and 
CAHG of P'; for each of these spaces must be repeated the 
same number of times,to complete the polygons to which they 
severally belong. 

First. The triangles ACD, ACG, whose common vertex is 
A, are to each other as their bases CD, CG; they are also to 
each other as the polygons p and p f ; hence 
p :p r : : CD : CG. 

Again, the triangles CGA, CGE, whose common vertex is 
G, are to each other as their bases CA, CE ; they are also to 
each other as the polygons p f and P; hence 
pt : P : : CA : CE. 

But since AD is parallel to EG, we have CD : CG : : CA 
: CE; therefore, 

p ;: p r : : p' : P; 

that is, the polygon p' is a mean proportional between the 
polygons p and P. 

Secondly. The triangles CGH, CHE, having the common 
altitude CG, are to each other as their bases GH, HE. But 
since CH bisects the angle GCE, we have (Prop. XVII., B. 

IV.), 

GH : HE : : CG : CE : : CD : CA, or CG : : p : pf. 
Therefore, CGH : CHE : :p :p r ; 








106 


GEOMETRY. 


hence (Prop. VI., B. II.) 

CGH : CGH + CHE, or CGE : : p : p+p 
or 2CGH : CGE : : 2 p : p-\-p f . 

But 2CGH, or CGHA : CGE : : P' : P. 


2»P 

Therefore, P' : P : : 2 p : p+p f ; whence W— p ; 

that is, the polygon P' is found by dividing twice the product 
of P and p by the sum of p and p ; . 

Hence, by means of the polygons p and P, it is easy to find 
the polygons p f and P' having double the number of sides. 


PROPOSITION X. THEOREM. 

A circle being given , two similar polygons can always be 
found , the one described about the circle, and the other inscribed 
in it , which shall differ from each other by less than any as¬ 
signable surface. 

Let ACD be the given circle, and the square of X any 
given surface; a polygon can be inscribed in the circle 
ACD, and a similar polygon -be described about it, such 
that the difference between them shall be less than the 
square of X. 

Bisect AC a fourth part of the circumference, then bisect 
the half of this fourth, and so continue the bisection, until an 
arc is found whose chord AB is less than X. As this arc 
must be contained a certain number of times exactly in the 
whole circumference, if we apply chords AB, BC, &c., each 
equal to AB, the last will terminate at A, and a regular 
polygon ABCD, &c., will be inscribed in the circle. 

Next describe a similar polygon about the circle (Prop. 
VI.) ; the difference of these two polygons will be less than 
the square of X. 

Find the center G, and draw the 
diameter AD. Let EF be a side 
of the circumscribed polygon; and 
join EG, FG. These lines will pass 
through the points A and B, as was E 
shown in Prop. VI. Draw GH to 
the point of contact H ; it will bisect 
AB in I, and be perpendicular to it 
(Prop. VI., Sch., B. III.). Join, also, 

BD. 

Let P represent the circumscribed polygon, and p the in¬ 
scribed polygon. Then, because the polygons are similar, 
they are as the squares of the homologous sides EF and AB 





BOOK VI. 


107 


(Prop. XXVI., B. IV.); that is, because the triangles EFG, 
ABG are similar, as the square of EG to the square of AG, 
that is, of HG. 

Again, the triangles EHG, ABD, having their sides paral¬ 
lel to each other, are similar; and, therefore, 

EG : HG : : AD : BD. 

But the polygon P is to the polygon p as the square of EG 

to the square of HG; 

hence P : p : : AD 2 : BD 2 , 

and, by division, P : P— p :: AD 2 : AD 2 —BD S , or AB 2 . 

But the square of AD is greater than a regular polygon of 
eight sides described about the circle, because it contains 
that polygon; and for the same reason, the polygon of eight 
sides is greater than the polygon of sixteen, and so on. 
Therefore P is less than the square of AD ; and, consequent¬ 
ly (Def. 2, B. II.), P— p is less than the square of AB ; that is, 
less than the given square on X. Hence, the difference of 
the two polygons is less than the given surface. 

Cor. Since the circle can not be less than any inscribed 
polygon, nor greater than any circumscribed one, it follows 
that a polygon may be inscribed in a circle , and another de¬ 
scribed about it, each of which shall differ from the circle by 
less than any assignable surface. 


PROPOSITION XI. PROBLEM. 

To find the area of a circle whose radius is unity. 

If the radius of a circle be unity, the diameter will be rep¬ 
resented by 2, and the area of the circumscribed square will 
be 4; while that of the inscribed square, being half the cir¬ 
cumscribed, is 2. Now, according to Prop. IX., the surface 
of the inscribed octagon, is a mean proportional between the 
two squares p and P, so that = ^8=2.82843. Also, the 

2oP 16 

circumscribed octagon P r =—-—-=—--=3.31371. Hav- 

p-\-p' 2+y/S 

ing thus obtained the inscribed and circumscribed octagons, 
we may in the same way determine the polygons having 
twice the number of sides. We must put p = 2.82843, and 
P = 3.31371, and we shall have p r = Vpl* = 3.06147 ; and 

pr = ^Z_= 3 . 18260 . These polygons of 16 sides will furnish 
p+p f 

us those of 32; and thus we may proceed, until there is no 
difference between the inscribed and circumscribed polygons, 
at least for any number of decimal places which may be de- 





108 


GEOMETRY. 


sired. The following table gives the results of this computa¬ 
tion for five decimal places : 


Number of Sides. 

Inscribed Polygon. 

Circumscribed Polygon. 

4 

2.00000 

4.00000 

8 

2.82843 

3.31371 

16 

3.06147 

3.18260 

32 

3.12145 

3.15172 

64 

3.13655 

3.14412 

128 

3.14033 

3.14222 

256 

3.14128 

3.14175 

512 

3.14151 

3.14163 

1024 

3.14157 

3.14160 

2048 

3.14159 

3.14159 


Now as the inscribed polygon can not be greater than the 
circle, and the circumscribed polygon can not be less than 
the circle, it is plain that 3.14159 must express the area of a 
circle, whose radius is unity, correct to five decimal places. 

After three bisections of a quadrant of a circle, we obtain 
the inscribed polygon of 32 sides, which differs from the cor¬ 
responding circumscribed polygon, only in the second decimal 
place. After five bisections, we obtain polygons of 128 sides, 
which differ only in the third decimal place ; after nine bisec¬ 
tions, they agree to five decimal places, but diffey in the sixth 
place; after eighteen bisections, they agree to ten decimal 
places ; and thus, by continually bisecting the arcs which sub¬ 
tend the sides of the polygon, new polygons are formed, both 
inscribed and circumscribed, which agree to a greater num¬ 
ber of decimal places. Vieta, by means of inscribed and cir¬ 
cumscribed polygons, carried the approximation to ten places 
of figures; Van Ceuleil carried it to 36 places; Sharp com¬ 
puted the area to 72 places; De Lagny to 128 places; and 
Rutherford has carried the computation to 208 places of de¬ 
cimals. 

By continuing this process of bisection, the difference be¬ 
tween the inscribed and circumscribed polygons may be 
made less than any quantity we can assign, however small. 
The number of sides of such a polygon will be indefinitely 
great; and hence a regular polygon of an infinite number of 
sides, is said to be ultimately equal to the circle. Henceforth, 
we shall therefore regard the circle as a regular polygon of 
an infinite number of sides. 


BOOK VI. 


109 


PROPOSITION XII. THEOREM. 



The area of a circle is equal to the product of its circum¬ 
ference by half the radius . 

Let ABE be a circle whose center is C 
and radius CA; the area of the circle is 
equal to the product of its circumference by 
half of CA. 

Inscribe in the circle any regular polygon, 
and from the center draw CD perpendicular 
to one of the sides. The area of the poly¬ 
gon will be equal to its perimeter multiplied 
by half of CD (Prop. VII.). Conceive the number of sides 
of the polygon to be indefinitely increased, by continually 
bisecting the arcs which subtend the sides; its perimeter 
will ultimately coincide with the circumference of the circle, 
the perpendicular CD will become equal to the radius CA, 
and the area of the polygon to the area of the circle (Prop. 
XI.). Consequently, the area of the circle is equal to the 
product of its circumference by half the radius. 

Cor . The area of a sector is equal to the product of its arc 
by half its radius . 

For the sector ACB is to the whole circle 
ABD, as the arc AEB is to the whole cir¬ 
cumference ABD (Prop. XIV., Cor. 2, B. 

III.); or, since magnitudes have the same 
ratio which their equimultiples have (Prop. 

VIII., B. II.), as the arc AEBx^AC is to the 

circumference ABD X |AC. But this last ex- V._ 

pression is equal to the area of the circle; 

therefore the area of the sector ACB is equal to the product 

of its arc AEB by half of AC. 



PROPOSITION XIII. THEOREM. 

The circumferences of circles are to each other as their radii t 
and their areas are as the squares of their radii . 

Let R and r denote the radii of two circles; C and c their 
circumferences ; A and a their areas ; then we shall have 
C : c : : R : r, 
and A : a : : R 2 : r 9 . 

Inscribe within the circles, two regular polygons having 




110 


GEOMETRY. 


the same number of sides. Now whatever be the number 
of sides of the polygons, their perimeters will be to each other 
as the radii of the circumscribed circles (Prop. VIII.). Con¬ 
ceive the arcs subtending the sides of the polygons to be con¬ 
tinually bisected, until the number of sides of the polygons 
becomes indefinitely great, the perimeters of the polygons 
will ultimately become equal to the circumferences of the 
circles, and we shall have 

C : c : : R : r. 

Again, the areas of the polygons are to each other as the 
squares of the radii of the circumscribed circles (Prop. VIII.). 
But when the number of sides of the polygons is indefinitely 
increased, the areas of the polygons become equal to the 
areas of the circles, and we shall have 
A : : : R 2 : r 3 . 

Cor. 1. Similar arcs are to each other as their radii; and 
similar sectors are as the squares of their radii. 

For since the arcs AB, ab are _ 

similar, the angle C is equal to the ^ 'y a _7, 

angle c (Def. 5, B. IV.). But the \ / \ / 

angle C is to four right angles, as \ \ / 

the arc AB is to the whole circum- \ / \/ 

ference described with the radius V c 

AC (Prop. XIV., B. III.); and the c 
angle c is to four right angles, as the arc ab is to the circum¬ 
ference described with the radius ac. Therefore the arcs 

AB, ab are to each other as the circumferences of which they 
form a part. But these circumferences are to each other as 

AC, ac; therefore, 

Arc AB : arc ab : : AC : ac. 


For the same reason, the sectors ACB, acb are as the en¬ 
tire circles to which they belong; and these are as the squares 
of their radii; therefore, 

Sector ACB : sector acb : : AC 2 : ac 3 . 

Cor. 2. Let n represent the circumference of a circle whose 
diameter is unity; also, let D represent the diameter, II the 
radius, and C the circumference of any other circle; then, 
since the circumferences of circles are to each other as their 
diameters, 


1 : u : : 2R : C; 
therefore, C = 2ttR = rrD ; 

that is, the circumference of a circle is equal to the product of 
its diameter by the constant number t r. 

Cor. 3. According to Prop. XII., the area of a circle is 
equal to the product of its circumference by half the radius. 
If we put A to represent the area of a circle, then 
A = C X |R = 27rR X iR = 7rR 2 ; 


BOOK VI. 


Ill 


that is, the area of a circle is equal to the product of the square 
of its radius by the constant number n. 

Cor. 4. When R is equal to unity, we have A=7r; that is, 
rc is equal to the area of a circle whose radius is unity. Ac¬ 
cording to Prop. XI., n is therefore equal to 3.14159 nearly. 
This number is represented by n, because it is the first letter 
of the Greek word which signifies circumference. 


112 


GEOMETRY. 


SOLID GEOMETRY. 


BOOK VII. 

PLANES AND SOLID ANGLES. 

Definitions . 

1. A straight line is perpendicular to a 
plane , when it is perpendicular to every 
straight line which it meets in that plane. 

Conversely, the plane in this case is per¬ 
pendicular to the line. 

The foot of the perpendicular, is the 
point in which it meets the plane. 

2. A line is parallel to a plane , when it can not meet the 
plane, though produced ever so far. 

Conversely, the plane in this case is parallel to the line. 

3. Two planes are parallel to each other, when they can 
not meet, though produced ever so far. 

4. The angle contained by two planes which cut each other, 
is the angle contained by two lines drawn 
from any point in the line of their common 
section, at right angles to that line, one in 
each of the planes. 

This angle may be acute, right, or obtuse. 

If it is a right angle, the two planes are 
perpendicular to each other. 

5. A solid angle is the angular space c 
tained by more than two planes which mee 
the same point. 




PROPOSITION I. THEOREM. 

One part of a straight line can not be in a plane, and another 
part without it. 

For from the definition of a plane (Def. 6, B. I.), when a 









BOOK VII. 


113 


straight line has two points common with a plane, it lies 
wholly in that plane. 

Scholium. To discover whether a surface is plane, we ap¬ 
ply a straight line in different directions to this surface, and 
see if it touches throughout its whole extent. 


PROPOSITION II. THEOREM. 

Any two straight lines which cut each other , are in one plane , 
and determine its position . 

Let the two straight lines AB, BC cut 
each other in B; then will AB, BC be in 
the same plane. 

Conceive a plane to pass through the 

straight line BC, and let this plane be turned _ 

about BC, until it pass through the point A. ® c 

Then, because the points A and B are situated in this plane, 
the straight line AB lies in it (Def. 6, B. I.). Hence the posi¬ 
tion of the plane is determined by the condition of its con¬ 
taining the two lines AB, BC. Therefore, any two straight 
lines, &c. 

Cor. 1 . A triangle ABC, or three points A, B, C, not in the 
same straight line, determine the position of a plane. 

Cor. 2. Two parallel lines AB, CD N 

determine the position of a plane. For ^B 
if the line EF be drawn, the plane of \ 

the two straight lines AE, EF will be C-v-D 

the same as that of the parallels AB, 

CD; and it has already been proved that two straight lines 
which cut each other, determine the position of a plane. 



PROPOSITION III. THEOREM. 


If two planes cut each other , their common section is a 
straight line. 


Let the two planes AB, CD cut each 
other, and let E, F be two points in their 
common section. From E to F draw the 
straight line EF. Then, since the points E 
and F are in the plane AB, the straight line 
EF which joins them, must lie wholly in 
that plane (Def. 6, B. I.). For the same 
reason, EF must lie wholly in the plane 

H 































114 


GEOMETRY. 


CD. Therefore the straight line EF is common to the two 
planes AB, CD; that is, it is their common section. Hence, 
if two planes, &c. 


PROPOSITION IV. THEOREM. 

If a straight line be 'perpendicular to each of two straight 
lines at their point of intersection , it will be perpendicular to 
the plane in which these lines are. 

Let the straight line AB be perpen¬ 
dicular to each of the straight lines 
CD, EF which intersect at B; AB will 
also be perpendicular to the plane MN 
which passes through these lines. 

Through B draw any line BG, in the 
plane MN; let G be any point of this 
line, and through G draw DGF, so that 
DG shall be equal to GF (Prob. XXI., 

B. V.). Join AD, AG, and AF. 

Then, since the base DF of the triangle DBF is bisected 
in G, we shall have (Prop. XIV., B. IV.), 
BD 2 +BF 2 =2BG 3 +2GF 2 . 

Also, in the triangle DAF, 

AD 2 +AF 2 =2AG 2 +2GF 2 . 

Subtracting the first equation from the second, we have 
AD 2 - BD 2 +A F 2 - BF 2 =2 AG 2 - 2BG\ 

But, because ABD is a right-angled triangle, 

AD 2 —BD 2 =AB 2 ; 

and, because ABF is a right-angled triangle, 

AF 2 —BF 2 =AB 2 . 

Therefore, substituting these values in the former equation, 
AB 2 -fAB 2 = 2AG 2 — 2BG 2 ; 
whence AB 2 = AG 2 —BG 9 , 

or AG 2 =AB 2 +BG 2 . 

Wherefore ABG is a right angle (Prop. XIII., Sch., B. IV.); 
that is, AB is perpendicular to the straight line BG. In like 
manner, it may be proved that AB is perpendicular to any 
other straight line passing through B in the plane MN; hence 
it is perpendicular to the plane MN (Def. 1). Therefore, if 
a straight line, &c. 

Scholium. Hence it appears not only that a straight line 
map be perpendicular to every straight line which passes 
through its foot in a plane, but that it always must be so, 
whenever it is perpendicular to two lines in the plane, which 
shows that the first definition involves no impossibility. 








BOOK VII. 


115 


Cor. 1 . The perpendicular AB is shorter than any oblique 
line AD ; it therefore measures the true distance of the point 
A from the plane MN. 

Cor. 2. Through a given point B in a plane, only one per¬ 
pendicular can be drawn to this plane. For, if there could 
be two perpendiculars, suppose a plane to pass through them, 
whose intersection with the plane MN is BG; then these 
two perpendiculars would both be at right angles to the line 
BG, at the same point and in the same plane, which is im¬ 
possible (Prop. XVI., Cor., B. I.). 

It is also impossible, from a given point without a plane, to 
let fall two perpendiculars upon the plane. For, suppose AB, 
AG to be two such perpendiculars; then the triangle ABG 
will have two right angles, which is impossible (Prop. XXVII., 
Cor. 3, B. I.). 


PROPOSITION V. THEOREM. 

Oblique lines drawn from a point to a plane , at equal dis¬ 
tances from the perpendicular , are equal; and of two oblique 
lines unequally distant from the perpendicular, the more remote 
is the longer . 

Let the straight line AB be 
drawn perpendicular to the plane 
MN; and let AC, AD, AE be ob¬ 
lique lines drawn from the point A, 
equally distant from the perpendic¬ 
ular ; also, let AF be more remote 
from the perpendicular than AE; 
then will the lines AC, AD, AE all 
be equal to each other, and AF be 
longer than AE. 

For, since the angles ABC, ABD, ABE are right angles, 
and BC, BD, BE are equal, the triangles ABC, ABD, ABE 
have two sides and the included angle equal; therefore the 
third sides AC, AD, AE are equal to each other. 

So, also, since the distance BF is greater than BE. it is 
plain that the oblique line AF is longer than AE (Prop. XVII., 
B. I.). 

Cor. All the equal oblique lines AC, AD, AE, &c., termi¬ 
nate in the circumference CDE, which is described from B, 
the foot of the perpendicular, as a center. 

If, then, it is required to draw a straight line perpendicular 
to the plane MN, from a point A without it, take three points 
in the plane C, D, E, equally distant from A, and find B the 






116 


GEOMETRY. 


center of the circle which passes through these points. Join 
AB, and it will be the perpendicular required. 

Scholium. The angle AEB is called the inclination of the 
line AE to the plane MN. All the lines AC, AD, AE,\fcc., 
which are equally distant from the perpendicular, have the 
same inclination to the plane; because all the angles ACB, 
ADB, AEB, &c., are equal. 


PROPOSITION VI. THEOREM. 

If a straight line is perpendicular to a plane , every plane 
which passes through that line , is perpendicular to the first- 
mentioned plane. 

Let the straight line AB be perpen¬ 
dicular to the plane MN; then will 
every plane which passes through AB 
be perpendicular to the plane MN. 

Suppose any plane, as AE, to pass 
through AB, and let EF be the common 
section of the planes AE, MN. In the 
plane MN, through the point B, draw 
CD perpendicular to the common sec¬ 
tion EF. Then, since the line AB is perpendicular to the 
plane MN, it must be perpendicular to each of the two 
straight lines CD, EF (Def. 1). But the angle ABD, formed 
by the two perpendiculars BA, BD, to the common section 
EF, measures the angle of the two planes AE, MN (Def. 4); 
and since this is a right angle, the two planes must be per¬ 
pendicular to each other. Therefore, if a straight line, &c. 

Scholium. When three straight lines, as AB, CD, EF, are 
perpendicular to each other, each of these lines is perpen¬ 
dicular to the plane of the other two, and the three planes 
are perpendicular to each other. 



PROPOSITION VII. THEOREM. 

If two planes are perpendicular to each other , a straight line 
drawn in one of them perpendicular to their common section , 
will he perpendicular to the other plane. 

Let the plane AE be perpendicular to the plane MN, and 
let the line AB be drawn in tho plane AE perpendicular to 
the common section EF; then will AB be perpendicular to 
the plane MN. 









BOOK vir. 


117 


For in the plane MN, draw CD 
through the point B perpendicular to 
EF. Then, because the planes AE and 
MN are perpendicular, the angle ABD 
is a right angle. Hence the line AB is 
perpendicular to the two straight lines 
CD, EF at their point of intersection; 
it is consequently perpendicular to their 
plane MN (Prop. IV.). Therefore, if 
two planes, &c. 

Cor. If the plane AE is perpendicular to the plane MN, 
and if from any point B, in their common section, we erect a 
perpendicular to the plane MN, this perpendicular will be in 
the plane AE. For if not, then we may draw from the same 
point, a straight line AB in the plane AE perpendicular to 
EF, and this line, according to the Proposition, will be per¬ 
pendicular to the plane MN. Therefore there would be two 
perpendiculars to the plane MN, drawn from the same point, 
which is impossible (Prop. IV., Cor. 2). 



PROPOSITION VIII. THEOREM. 

If two planes , which cut one another , are each of them per - 
pendicular to a third plane , their common section is perpen¬ 
dicular to the same plane. 

Let the two planes AE, AD be each 
of them perpendicular to a third plane 
MN, and let AB be the common sec¬ 
tion of the first two planes; then will 
AB be perpendicular to the plane MN. 

For, from the point B, erect a per¬ 
pendicular to the plane MN. Then, by 
the Corollary of the last Proposition, 
this line must be situated both in the 
plane AD and in the plane AE; hence it is their common 
section AB. Therefore, if two planes, &c. 



PROPOSITION IX. THEOREM. 

% 

Two straight lines which are perpendicular to the same plane, 
are parallel to each other . 

Let the two straight lines AB, CD be each of them perpen¬ 
dicular to the same plane MN; then will AB be parallel to CD. 



























118 


GEOMETRY. 


In the plane MN, draw the straight 
line BD joining the points B and D. 

Through the lines AB, BD pass the 
plane EF; it will be perpendicular to 
the plane MN (Prop. VI.); also, the 
line CD will lie in this plane, because it 
is perpendicular to MN (Prop. VII., 

Cor.). Now, because AB and CD are 
both perpendicular to the plane MN, 
they are perpendicular to the line BD in that plane ; and since 
AB, CD are both perpendicular to the same line BD, and lie 
in the same plane, they are parallel to each other (Prop. 
XX., B. I.). Therefore, two straight lines, &c. 

Cor. 1 . If one of two parallel lines be perpendicular to a 
plane, the other will be perpendicular to the same plane. If 
AB is perpendicular to the plane MN, then (Prop. VI.) the 
plane EF will be perpendicular to MN. Also, AB is per¬ 
pendicular to BD; and if CD is parallel to AB, it will be 
perpendicular to BD, and therefore (Prop. VII.). it is perpen¬ 
dicular to the plane MN. 

Cor. 2. Two straight lines, parallel to a third, are parallel 
to each other. For, suppose a plane to be drawn perpen¬ 
dicular to any one of them; then the other two, being paral¬ 
lel to the first, will be perpendicular to the same plane, by 
the preceding Corollary; hence, by the Proposition, they will 
be parallel to each other. 

The three straight lines are supposed not to be in the same 
plane; for in this case the Proposition has been already de¬ 
monstrated. 



PROPOSITION X. THEOREM. 

If a straight line , without a given plane , he parallel to a 
straight line in the plane , it will he parallel to the plane . 

Let the straight line AB be parallel 
to the straight line CD, in the plane 
MN; then will it be parallel to the 
plane MN. 

Through the parallels AB, CD sup¬ 
pose a plane ABDC to pass. If the line 
AB can meet the plane MN, it must 
meet it in some point of the line CD, which is the common 
intersection of the two planes. But AB can not meet CD, 
since they are parallel; hence it can not meet the plane MN; 
that is, AB is parallel to the plane MN (Def. 2). Therefore, 
if a straight line, &c. 


















BOOK VII. 


119 


PROPOSITION XI. THEOREM. 

Two planes , which are perpendicular to the same straight 
line , are parallel to each other. 

Let the planes MN, PQ be 
perpendicular to the line AB; 
then will they be parallel to each 
other. m 

For if they are not parallel, 
they will meet if produced. Let 
them be produced and meet in C. 

Join AC, BC. Now the line AB, 
which is perpendicular to the plane MN, is perpendicular to 
the line AC drawn through its foot in that plane. For the 
same reason AB is perpendicular to BC. Therefore CA and 
CB are two perpendiculars let fall from the same point C 
upon the same straight line AB, which is impossible (Prop. 
NVI., B. I.). Hence the planes MN, PQ, can not meet when 
produced; that is, they are parallel to each other. There¬ 
fore, two planes, &c. 



PROPOSITION XII. THEOREM. 


If two parallel planes are cut by a third plane , their common 
sections are parallel. 

Let the parallel planes MN, PQ be p 

cut by the plane ABDC; and let their 
common sections with it be AB, CD; 
then will AB be parallel to CD.* 

For the two lines AB, CD are in the 
same plane, viz., in the plane ABDC 
which cuts the planes MN, PQ ; and 
if these lines wei«e not parallel, they 
would meet when produced; therefore 
the planes MN, PQ would also meet, which is impossible, be¬ 
cause they are parallel. Hence the lines AB, CD are paral¬ 
lel. Therefore, if two parallel planes, &c. 













































120 


GEOMETRY. 


PROPOSITION XIII. THEOREM. 



If two planes are parallel , a straight line which is perpen¬ 
dicular to one of them, is also perpendicular to the other. 

Let the two planes MN, PQ be par¬ 
allel, and let the straight line AB be 
perpendicular to the plane MN; AB ^ 
will also be perpendicular to the plane 
PQ. 

Through the point B, draw any line 
BD in the plane PQ; and through the P 
lines AB, BD suppose a plane to pass intersecting the plane 
MN in AC. The two lines AC, BD will be parallel (Prop. 
XII.). But the line AB, being perpendicular to the plane 
MN, is perpendicular to the straight line AC which it meets 
in that plane ; it must, therefore, be perpendicular to its par¬ 
allel BD (Prop. XXIII., Cor. 1, B. I.). But BD is any line 
drawn through B in the plane PQ; and since AB is perpen¬ 
dicular to any line drawn through its foot in the plane PQ, 
it must be perpendicular to the plane PQ (Def. 1). There¬ 
fore, if two planes, &c. 


PROPOSITION XIV. THEOREM. 

Parallel straight lines included between two parallel planes 
are equal. 

Let AB, CD be the two parallel 
straight lines included between two 
parallel planes MN, PQ; then will AB 
be equal to CD. 

Through the two parallel lines AB, 

CD suppose a plane ABDC to pass, in¬ 
tersecting the parallel planes in AC and 
BD. The lines AC, BD will be parallel to each other (Prop. 
XII.). But AB is, by supposition, parallel to CD ; therefore 
the figure ABDC is a parallelogram; and, consequently, AB 
is equal to CD (Prop. XXIX., B. I.). Therefore, parallel 
straight lines, &c. 

Cor. Hence two parallel planes are every where equidis¬ 
tant; for if AB, CD are perpendicular to the plane MN, they 
will be perpendicular to the parallel plane PQ (Prop. XIII.); 
and being both perpendicular to the same plane, they will be 
parallel to each other (Prop. IX.), and, consequently, equal. 











BOOK VII. 


121 



PROPOSITION XV. THEOREM. 

If two angles, not in the same plane, have their sides paral¬ 
lel and similarly situated, these angles will he equal, and their 
planes will he parallel. 

Let the two angles ABC, DEF, 
lying in different planes MN, PQ, ^ 
have their sides parallel each to each 
and similarly situated; then will the 
angle ABC be equal to the angle 
DEF, and the plane MN be parallel 
to the plane PQ. 

Take AB equal to DE, and BC 
equal to EF, and join AD, BE, CF, 

AC, DF. Then, because AB is equal and parallel to DE, 
the figure ABED is a parallelogram (Prop. XXXI., B. I.); 
and AD is equal and parallel to BE. For the same reason 
CF is equal and parallel to BE. Consequently, AD and CF, 
being each of them equal and parallel to BE,, are parallel to 
each other (Prop. IX., Cor. 2), and also equal; therefore AC 
is also equal and parallel to DF (Prop. XXXI., B. I.). Hence 
the triangles ABC, DEF are mutually equilateral, and the 
angle ABC is equal to the angle DEF (Prop. XV., B. I.). 

Also, the plane ABC is parallel to the plane DEF. For, if 
they are not parallel, suppose a plane to pass through A par¬ 
allel to DEF, and let it meet the straight lines BE, CF in the 
points G and H. Then the three lines AD, GE, HF will be 
equal (Prop. XIV.). But the three lines AD, BE, CF have 
already been proved to be equal; hence BE is equal to GE, 
and CF is equal to HF, which is absurd ; consequently, the 
plane ABC must be parallel to the plane DEF. Therefore, 
if two angles, &c. 

Cor. 1 . If two parallel planes MN, PQ are met by two 
other planes ABED, BCFE, the angles formed by the inter¬ 
sections of the parallel planes will be equal. For the section 
AB is parallel to the section DE (Prop. XII.); and BC is 
parallel to DF; therefore, by the Proposition, the angle ABC 
is equal to the angle DEF. 

Cor. 2. If three straight lines AD, BE, CF, not situated in 
the same plane, are equal and parallel, the triangles ABC, 
DEF, formed by joining the extremities of these lines, will be 
equal, and their planes will be parallel. For, since AD is 
equal and parallel to BE, the figure ABED is a parallel¬ 
ogram ; hence the side AB is equal and parallel to DE. Foi 






122 


GEOMETRY. 


the same reason, the sides BC and EF are equal and paral¬ 
lel; as, also, the sides AC and DF. Consequently, the two 
triangles ABC, DEF are equal; and, according to the Prop¬ 
osition, their planes are parallel. 


PROPOSITION XVI. THEOREM. 



If two straight lines are cut by parallel planes , they will be 
cut in the same ratio. 

Let the straight lines AB, CD be cut 
by the parallel planes MN, PQ, RS in 
the points A, E, B, C, F, D; then we 
shall have the proportion 

AE : EB : : CF : FD. 

Draw the line BC meeting the plane 
PQ in G, and join AC, BD, EG, GF. 

Then, because the two parallel planes 
MN, PQ are cut by the plane ABC, the ^ S ■ 

common sections AC, EG are parallel (Prop. XII.). Also, be¬ 
cause the two parallel planes PQ, RS are cut by the plane 
BCD, the common sections BD, GF are parallel. Now, be¬ 
cause EG is parallel to AC, a side of the triangle ABC (Prop. 
XVI., B. IV.), we have 

AE : EB : : CG : GB. 

Also, because GF is parallel to BD, one side of the triangle 
BCD, we have 

CG : GB : : CF : FD; 
hence (Prop. IV., B. II.), 

AE : EB : : CF : FD. 

Therefore, if two straight lines, &c. 


PROPOSITION XVII. THEOREM. 

If a solid angle is contained by three plane angles , the sum 
of any two of these angles is greater than the third. 

Let the solid angle at A be con¬ 
tained by the three plane angles 
BAC, CAD, DAB; any two of these 
angles will be greater than the third. 

If these three angles are all equal 
to each other, it is plain that any 
two of them must be greater than 
the third. But if they are not equal, 









BOOK VII. 


123 


let BAC be that angle which is not less than either of the 
other two, and is greater than one of them BAD. Then, at 
the point A, make the angle BAE equal to the angle BAD ; 
take AE equal to AD; through E draw the line BEC cutting 
AB, AC in the points B and C ; and join DB, DC. 

Now, because, in the two triangles BAD, BAE, AD is 
equal to AE, AB is common to both, and the angle BAD is 
equal to the angle BAE; therefore the base BD is equal to 
the base BE (Prop. VI., B. I.). Also, because the sum of the 
lines BD, DC is greater than BC (Prop. VIII., B. I.), and BD 
is proved equal to BE, a part of BC, therefore the remaining 
line DC is greater than EC. Now, in the two triangles 
CAD, CAE, because AD is equal to AE, AC is common, but 
the base CD is greater than the base CE; therefore the an¬ 
gle CAD is greater than the angle CAE (Prop. XIV., B. I.). 
But, by construction, the angle BAD is equal to the angle 
BAE; therefore the two angles BAD, CAD are together 
greater than BAE, CAE; that is, than the angle BAC. 
Now BAC is not less than either of the angles BAD, CAD ; 
hence BAC, with either of them, is greater than the third. 
Therefore, if a solid angle, &c. 


PROPOSITION XVIII. THEOREM. 

The plane angles which contain any solid angle , are together 
less than four right angles. 

Let A be a solid angle contained by any 
number of plane angles BAC, CAD, DAE, 

EAF, FAB; these angles are together less 
than four right angles. 

Let the planes which contain the solid an¬ 
gle at A be cut by another plane, forming 
the polygon BCDEF. Now, because the 
solid angle at B is contained by three plane 
angles, any two of which are greater than 
the third (Prop. XVII.), the two angles ABC, 

ABF are greater than the angle FBC. For 
the same reason, the two angles ACB, ACD are greater than 
the angle BCD, and so with the other angles of the polygon 
BCDEF. Hence, the sum of all the angles at the bases of the 
triangles having the common vertex A, is greater than the 
sum of all the angles of the polygon BCDEF. But all the 
angles of these triangles are together equal to twice as many 
right angles as there are triangles (Prop. XXVII., B. I.), that 
is^ as there are sides of the polygon BCDEF. Also, the an- 



124 


GEOMETRY. 


gles of the polygon, together with four right angles, are equal 
to twice as many right angles as the figure has sides (Prop. 
XXVIII., B. I.); hence all the angles of the triangles are 
equal to all the angles of the polygon, together with four 
right angles. But it has been proved that the angles at the 
bases of the triangles, are greater than the angles of the 
polygon. Hence the remaining angles of the triangles, viz., 
those which contain the solid angle at A, are less than four 
right angles. Therefore, the plane angles, &c. 

Scholium. This demonstration supposes that the solid an¬ 
gle is convex; that is, that the plane of neither of the faces, 
if produced, would cut the solid angle. If it were otherwise, 
the sum of the plane angles would no longer be limited, and 
might be of any magnitude. 


PROPOSITION XIX. THEOREM. 

If two solid angles are contained hy three plane angles which 
are equal, each to easily the planes of the equal angles will he 
equally inclined.to each other. 

Let A and a be two solid 
angles, contained by three 
plane angles which are 
equal, each to each, viz., the 
angle BAC equal to hac, 
the angle CAD to cad, and 
BAD equal to had ; then 
will the inclination of the 
planes ABC, ABD be equal 
to the inclination of the planes ahc, ahd. 

In the line AC, the common section of the planes ABC, 
ACD, take any point C ; and through C let a plane BCE 
pass perpendicular to AB, and another plane CDE perpen¬ 
dicular to AD. Also, take ac equal to AC-; and through c 
let a plane bee pass perpendicular to ah, and another plane 
ede perpendicular to ad. 

Now, since the line AB is perpendicular to the plane BCE, 
it is perpendicular to every straight line which it meets in 
that plane ; hence ABC and ABE are right angles. For the 
same reason ahc and ahe are right angles. Now, in the tri¬ 
angles ABC, ahc, the angle BAC is, by hypothesis, equal to 
hac, and the angles ABC, ahc are right angles; therefore the 
angles ACB, ach are equal. But the side AC was made 
equal to the side ac ; hence the two triangles are equal 
(Prop. VII., B. I.); that is, the side AB is equal to ah, and BC 





BOOK VII. 


125 


to be. In the same manner, it may be proved that AD is 
equal to ad , and CD to cd. 

We can now prove that the quadrilateral ABED is equal 
to the quadrilateral abed. For, let the angle BAD be placed 
upon the equal angle bad , then the point B will fall upon the 
point b , and the point D upon the point d; because AB is 
equal to ab, and AD to ad. At the same time, BE, which is 
perpendicular to AB, will fall upon be, which is perpendicu¬ 
lar to ab ; and for a similar reason DE will fall upon de. 
Hence the point E will fall upon e, and we shall have BE 
equal to be, and DE equal to de. 

Now, since the plane BCE is perpendicular to the line AB, 
it is perpendicular to the plane ABD which passes through 
AB (Prop. VI.). For the same reason CDE is perpendicular 
to the same plane; hence CE, their common section, is per¬ 
pendicular to the plane ABD (Prop. VIII.). In the same man¬ 
ner, it may be proved that ce is perpendicular to the plane 
abd. Now, in the triangles BCE, bee, the angles BEC, bee are 
right angles, the hypothenuse BC is equal to the hypothenuse 
be, and the side BE is equal to be ; hence the two triangles 
are equal, and the angle CBE is equal to the angle cbe. But 
the angle CBE is the inclination of the planes ABC, ABD 
(Def. 4) ; and the angle cbe is the inclination of the planes 
abc, abd; hence these planes are equally inclined to each 
other. 

We must, however, observe that the angle CBE is not, 
properly speaking, the inclination of the planes ABC, ABD, 
except when the perpendicular CE falls upon the same side 
of AB as AD does. If it fall upon the other side of AB, then 
the angle between the two planes will be obtuse, and this 
angle, together with the angle B of the triangle CBE, will 
make two right angles. But in this case, the angle between 
the two planes abc, abd will also be obtuse, and this angle, 
together with the angle b of the triangle cbe, will also make 
two right angles. And, since the angle A is always equal to 
the angle a, the inclination of the two planes ABC, ABD will 
always be equal to that of the planes abc, abd. Therefore, if 
two solid angles, &c. 

Scholium. If two solid angles are contained by three 
plane angles which are equal, each to each, and similarly 
situated, the angles will be equal, and will coincide when 
applied the one to the other. For we have proved that the 
quadrilateral ABED will coincide with its equal abed. 
Now, because the triangle BCE is equal to the triangle bee, 
the line CE, which is perpendicular to the plane ABED, is 
equal to the line ce, w r hich is perpendicular to the plane abed. 
And since only one perpendicular can be drawn to a plane 


126 


GEOMETRY. 


from the same point (Prop. 

IV., Cor. 2), the lines CE, ce 
must coincide with each oth¬ 
er, and the point C coincide 
with the point c. Hence 
the two solid angles must 
coincide throughout. B 

It should, however, be ob¬ 
served that the two solid an¬ 
gles do not admit of superposition, unless the three equal plane 
angles are similarly situated in both cases. For if the per¬ 
pendiculars CE, ce lay on opposite sides of the planes ABED, 
abed, the two solid angles could not be made to coincide. 
Nevertheless, the Proposition will always hold true, that the 
planes containing the equal angles are equally inclined to 
each other. 




BOOK VIII. 


127 


BOOK VIII. 

POLYEDRONS. 


Definitions . 


1. A polyedron is a solid included by any number of planes, 
which are called its faces. If the solid have only four faces, 
which is the least number possible, it is called a tetraedron; 
if six faces, it is called a hexaedron; if eight, an octaedron; 
if twelve, a dodecaedron ; if twenty, an icosaedron , &c. 

2. The intersections of the faces of a polyedron are called 
its edges. A diagonal of a polyedron is the straight line 
which joins any two vertices not lying in the same face. 

3. Similar polyedrons are such as have all their solid an¬ 
gles equal, each to each, and are contained by the same num¬ 
ber of similar planes. 

4. A regular polyedron is one whose solid angles are all 
equal to each other, and whose faces are all equal and regu¬ 
lar polygons. 

5. A prism is a polyedron having two faces 
which are equal and parallel polygons ; and 
the others are parallelograms. The equal 
and parallel polygons are called the bases of 
the prism; the other faces taken together 
form the lateral or convex surface. The alti¬ 
tude of a prism is the perpendicular distance 
between its two bases. 

6. A right prism is one whose edges are 
all perpendicular to the bases. Any other prism is called an 
oblique prism. 

7. A prism is triangular , quadrangular , pentagonal , hex¬ 
agonal, &c., according as its base is a triangle, a quadri¬ 
lateral, a pentagon, a hexagon, &c. 

8. A parallelopiped is a prism whose 
bases are parallelograms. A right par¬ 
allelopiped is one whose faces are all rect¬ 
angles. 

9. A cube is a right parallelopiped bounded by six equal 
squares. 











128 


GEOMETRY. 


10. A pyramid is a polyedron contained by 
several triangular planes proceeding from the 
same point, and terminating in the sides of a 
polygon. This polygon is called the base of 
the pyramid ; and the point in which the planes 
meet, is the vertex. The triangular planes form 
the convex surface. 

11. The altitude of a pyramid is the perpen¬ 
dicular let fall from the vertex upon the plane 
of the base, produced if necessary. The slant height of a 
pyramid is a line drawn from the vertex, perpendicular to 
one side of the polygon which forms its base. 

12. A pyramid is triangular , quadrangular, &c., according 
as the base is a triangle, a quadrilateral, &c. 

13. A regular pyramid is one whose base is a regular poly¬ 
gon, and the perpendicular let’ fall from the vertex upon the 
base, passes through the center of the base. This perpendic¬ 
ular is called the axis of the pyramid. 

14. A frustrum of a pyramid is a portion of the solid next 
the base, cut off by a plane parallel to the base. The alti¬ 
tude of the frustrum is the perpendicular distance between 
the two parallel planes. 



PROPOSITION I. THEOREM. 


The convex surface of a right prism is equal to the pe¬ 
rimeter of its base multiplied by its altitude. 

Let ABCDE-K be a right prism; then will 
its convex surface be equal to the perimeter 
of the base AB+BC+CD+DE+EA multi¬ 
plied by its altitude AF. 

For the convex surface of the prism is 
equal to the sum of the parallelograms AG, 

BH, Cl, &lc. Now the area of the parallelo¬ 
gram AG is measured by the product of its 
base AB by its altitude AF (Prop. IV., Sch., 

B. IV.). The area of the parallelogram BH is measured by 
BC xBG; the area of Cl is measured by CD xCH, and so of 
the others. But the lines AF, BG, CH, &c., are all equal to 
each other (Prop. XIV., B. VII.), and each equal to the alti¬ 
tude of the prism. Also, the lines AB, BC, CD, &c., taken 
together, from the perimeter of the base of the prism. There¬ 
fore, the sum of these parallelograms, or the convex surface 
of the prism, is equal to the perimeter of its base, multiplied 
by its altitude. 






















BOOK VIII. 


129 


Cor. If two right prisms have the same altitude, their con¬ 
vex surfaces will be to each other as the perimeters of their 
bases. 


PROPOSITION II. THEOREM. 

In every prism , the sections formed by parallel planes are 
equal polygons. 

Let the prism LP be cut by the parallel 
planes AC, FH ; then will the sections ABC 
DE, FGHIK, be equal polygons. 

Since AB and FG are the intersections 
of two parallel planes, with a third plane 
LMON, they are parallel. The lines AF, 

BG are also parallel, being edges of the 
prism; therefore ABGF is a parallelogram, 
and AB is equal to FG. For the same iUff 

reason BC is equal and parallel to GH, CD 
to IH, DE to IK, and AE to FK. M 

Because the sides of the angle ABC are parallel to those of 
FGH, and are similarly situated, the angle ABC is equal to 
FGH (Prop. XV., B. VII.). In like manner, it may be proved 
that the angle BCD is equal to the angle GHI, and so of the 
rest. Therefore the polygons ABCDE, FGHIK are equal. 

Cor. Every section of a prism,made parallel to the base, is 
equal to the base. 



PROPOSITfON III. THEOREM. 

Two prisms are equal , when they have a solid angle contain - 
ed by three faces which are equal , each to each , and similarly 
situated. 

Let AT, ai be two prisms 
having the faces which 
contain the solid angle B 
equal to the faces which 
contain the solid angle b; 
viz., the base ABCDE to 
the base abcde, the paral- mmmm 
lelogram AG to the paral- 
lelogram ag , and the par- ~ 

allelogram BH to the par- 
allelogram bh ; then will the prism AI be equal to the prism ai. 




130 


GEOMETRY. 


Let the prism AI be 
applied to the prism ai, so 
that the equal bases AD 
and ad may coincide, the 
point A falling upon a, B 
upon b , and so on. And 
because the three plane 
angles which contain the 
solid angle B, are equal 
to the three plane angles 
which contain the solid angle b, and these planes are similarly 
situated, the solid angles B and b are equal (Prop. XIX., Sch., 
B. VII.). Hence the edge BG will coincide with its equal bg ; 
and the point G will coincide with the point g. Now, be¬ 
cause the parallelograms AG and ag are equal, the side GF 
will fall upon its equal gf; and for the same reason, GH will 
fall upong/h Hence the plane of the base FGHIK will coin¬ 
cide with the plane of the bas efghik (Prop. II., B. VII.). But, 
since the upper bases are equal to their corresponding lower 
bases, they are equal to each other; therefore the base FI will 
coincide throughout with fi; viz., HI with hi , IK with ik, and 
KF with kf; hence the prisms coincide throughout, and are 
equal to each other. Therefore, two prisms, &c. 

Cor. Two right prisms, which have equal bases and equal 
altitudes, are equal. 

For, since the side AB is equal to ab , and the altitude BG 
to bg, the rectangle ABGF is equal to the rectangle abgf. 
So, also, the rectangle BGHC is equal to the rectangle bghc ; 
hence the three faces which contain the solid angle B are 
equal to the three faces which contain the solid angle b; 
consequently, the two prisms are equal. 



PROPOSITION IV. THEOREM, 

The opposite faces of a par allelopiped are equal and parallel. 

Let ABGH be a parallelopiped; then 
will its opposite faces be equal and parallel. 

From the definition of a parallelopiped 
(Def. 8) the bases AC, EG are equal and 
parallel; and it remains to be proved that 
the same is true of any two opposite faces, 
as AH, BG. Now, because AC is a par¬ 
allelogram, the side AD is equal and par¬ 
allel to BC. For the same reason AE is equal and parallel 
to BF; hence the angle DAE is equal to the angle CBF 





BOOK VIII. 


131 


(Prop. XV., B. VII.), and the plane DAE is parallel to the 
plane CBF. Therefore also the parallelogram AH is equal 
to the parallelogram BG. In the same manner, it may be 
proved that the opposite faces AF and DG are equal and 
parallel. Therefore, the opposite faces, &c. 

Cor. 1 . Since a parallelopiped is a solid contained by six 
faces, of which the opposite ones are equal and parallel, any 
face may be assumed as the base of a parallelopiped. 

Cor. 2. The four diagonals of a parallelopiped bisect each 
other. 

Draw any two diagonals AG, EC; they G 

will bisect each other. Since AE is equal 
and parallel to CG, the figure AEGC is a 
parallelogram; and therefore the diago¬ 
nals AG, EC bisect each other (Prop. 

XXXII., B. I.). In the same manner, it 
may be proved that the two diagonals BH 
and DF bisect each other; and hence the 
four diagonals mutually bisect each other, in a point which 
may be regarded as the center of the parallelopiped. 



PROPOSITION V. THEOREM. 

If a parallelopiped be cut by a plane passing through the 
diagonals of two opposite faces , it will be divided into two 
equivalent prisms. 

Let AG be a parallelopiped, and AC, G 

EG the diagonals of the opposite parallelo¬ 
grams BD, FH. Now, because AE, CG are 
each of them parallel to BF, they are par- ° 
allel to each other ; therefore the diagonals 
AC, EG are in the same plane with AE, 

CG; and the plane AEGC divides the solid 
AG into two equivalent prisms. 

Through the vertices A and E draw the ^ 
planes AIKL, EMNO perpendicular to AE, 
meeting the other edges of the parallelo¬ 
piped in the points I, K, L, and in M, N, O. 

The sections AIKL, EMNO are equal, because they are 
formed by planes perpendicular to the same straight line, 
and, consequently, parallel (Prop. II.). They are also par¬ 
allelograms, because AI, KL, two opposite sides of the same 
section, are the intersections of two parallel planes ABFE, 
DCGH,by the same plane. 

For the same reason, the figure ALOE is a parallelogram ; 






132 


GEOMETRY. 


so, also, are AIME, IKNM, KLON, the 
other lateral faces of the solid AIKL- 
EMNO ; hence this solid is a prism (Def. 

5); and it is a right prism, because AE is 
the plane of its base. But 
AN is divided into two 
LK-N, AIK-N; for the 
bases of these prisms are equal, being 
halves of the same parallelogram AIKL, 
and they have the common altitude AE; 
they are therefore equal (Prop. III., Cor.). 

Now, because AEHD, AEOL are paral¬ 
lelograms, the sides DH, LO, being equal to AE, are equal to 
each other. Take away the common part DO, and we have 
DL equal to HO. For the same reason, CK is equal to GN. 
Conceive now that ENO, the base of the solid ENGHO, is 
placed on AKL, the base of the solid AKCDLr; then the 
point O falling on L and N on K, the lines HO, GN will coin¬ 
cide with their equals DL, CK, because they are perpendic¬ 
ulars to the same plane. Hence the two solids coincide 
throughout, and are equal to each other. To each of these 
equals, add the solid ADC-N; then will the oblique prism 
ADC-G be equivalent to the right prism A LK-N. 

In the same manner, it may be proved that the oblique 
prism ABC-G is equivalent to the right prism AIK-N. 
But the two right prisms have been proved to be equal; 
hence the two oblique prisms ADC-G, ABC-G are equiva¬ 
lent to each other. Therefore, if a parallelopiped, &c. 

Cor . Every triangular prism is half of a parallelopiped 
having the same solid angle, and the same edges AB, BC, 
BF. 

Scholium. The triangular prisms into which the oblique 
parallelopiped is divided, can not be made to coincide, be¬ 
cause the plane angles 'about the corresponding solid angles 
are not similarly situated. 


perpendicular to 
the right prism 
equal prisms A 



PROPOSITION VI. THEOREM. 

Parallelopipeds,of the same base and the same altitude f are 
equivalent. 

Case first. When their upper bases are between the same 
parallel lines. 

Let the parallelopipeds AG, AL have the base AC com¬ 
mon, and let their opposite bases EG, IL be in the same 
plane, and between the same parallels EK, HL ; then will 
the solid AG be equivalent to the solid AL. 



book virr. 


133 



Because AF, AK are parallel¬ 
ograms, EF and IK are each 
equal to AB, and therefore equal 
to each other. Hence, if EF and 
IK be taken away from the same 
line EK, the remainders El and 
FK will be equal. Therefore 
the triangle AEI is equal to the 
triangle BFK. Also, the parallelogram EM is equal to the 
parallelogram FL, and AH to BG. Hence the solid angles 
at E and F are contained by three faces which are equal to 
each other, and similarly situated ; therefore the prism AEI- 
M is equal to the prism BFK-L (Prop. III.). 

Now’, if from the whole solid AL, we take the prism 
AEI-M, there will remain the parallelopiped AL; and if 
from the same solid AL, we take the prism BFK-L, there 
will remain the parallelopiped AG. Hence the parallelopi- 
peds AL, AG are equivalent to one another. 

Case second. When their upper bases are not between the 
same parallel lines. 

Let the parallelopipeds AG, ^ 

AL have the same base AC and v™ 
the same altitude ; then will their 
opposite bases EG, IL be in the 
same plane. And, since the sides 
EF and IK are equal and parallel 
to AB, they are equal and paral¬ 
lel to each other. For the same 
reason FG is equal and parallel 
to KL. Produce the sides EH, 

EG, as also IK, LM, and let 
them meet in the points N, O, P, Q,; the figure NOPQ, is a 
parallelogram equal to each of the bases EG, IL; and, con¬ 
sequently, equal to ABCD, and parallel to it. 

Conceive now a third parallelopiped AP, having AC for its 
lower base, and NP for its upper base. The solid AP will 
be equivalent to the solid AG, by the first Case, because they 
have the same lower base, and their upper bases are in the 
same plane and between the same parallels, EQ, FP. For 
the same reason, the solid AP is equivalent to the solid AL; 
hence the solid AG is equivalent to the solid AL. There¬ 
fore, parallelopipeds, &c. 










134 


GEOMETRY. 


PROPOSITION VII. THEOREM. 

Any parallelopiped is equivalent to a right parallelopiped 
having the same altitude and an equivalent base. 

Let AL be any parallelopiped; it is equivalent to a right 
parallelopiped having the same altitude and an equivalent 

From the points A, B, C, D draw AE, BF, CG, DH, per¬ 
pendicular to the plane of the low¬ 
er base, meeting the plane of the 
upper base in the points E, F,.G, & 

H. Join EF, FG, GH, HE ; there 
will thus be formed the parallelo¬ 
piped AG, equivalent to AL (Prop. 

VI.); and its lateral faces AF, BG, 

CH, DE are rectangles. If the 
base ABCD is also a rectangle, 

AG will be a right parallelopiped, 
and it is equivalent to the parallel¬ 
opiped AL. But if ABCD is not a rectangle, from A and B 
draw AI, BK perpendicular to CD ; and ^ , r r 
from E and F draw EM, FL perpendicu¬ 
lar to GH ; and join IM, KL. The solid 
ABKI-M will be a right parallelopiped. 

For, by construction, the bases ABKI and 
EFLM are rectangles; so, also, are the 
lateral faces, because the edges AE, BF, 

KL, IM are perpendicular to the plane of 
the base. Therefore the solid AL is a right 
parallelopiped. But the two parallelopipeds ^ B 

AG, AL may be regarded as having the same base AF, and 
the same altitude AI; they are therefore equivalent. But the 
parallelopiped AG is equivalent to the first supposed parallel¬ 
opiped ; hence this parallelopiped is equivalent to the right 
parallelopiped AL, having the same altitude, and an equiva¬ 
lent base. Therefore, any parallelopiped, &c. 






BOOK VIII. 


135 


PROPOSITION VIII. THEOREM. 

Right parallelepipeds, having the same base, are to each oth¬ 
er as their altitudes. 

Let AG, AL be two right parallelopipeds 
having the same base ABCD ; then will they 
be to each other as their altitudes AE, AI. 

Case first. When the altitudes are in the 
ratio of two whole numbers. 

Suppose the altitudes AE, AI are in the 
ratio of two whole numbers ; for example, as 
seven to four. Divide AE into seven equal 
parts; AI will contain four of those parts. 

Through the several points of division, let 
planes be drawn parallel to the base; these 
planes will divide the solid AG into seven 
small parallelopipeds, all equal to each other, having equal 
bases and equal altitudes. The bases are equal, because ev¬ 
ery section of a prism parallel to the base is equal to the base 
(Prop. II., Cor.); the altitudes are equal, for these altitudes 
are the equal divisions of the edge AE. But of these seven 
equal parallelopipeds, AL contains four; hence the solid AG 
is to the solid AL, as seven to four, or as the altitude AE is 
to the altitude AI. 

Case second. When the altitudes are not in the ratio of two 
whole numbers. 

Let AG, AL be two parallelopipeds whose altitudes have 
any ratio whatever; we shall still have the proportion 
Solid AG : solid AL : : AE : AI. 

For if this proportion is not true, the first three terms re¬ 
maining the same, the fourth term must be greater or less 
than AI. Suppose it to be greater, and that we have 
Solid AG : solid AL : : AE : AO. 

Divide AE into equal parts each less than 01; there will 
be at least one point of division between 0 and I. Designate 
that point by N. Suppose a parallelopiped to be construct¬ 
ed, having ABCD for its base, and AN for its altitude; and 
represent this parallelopiped by P. Then, because the alti¬ 
tudes AE, AN are in the ratio of two whole numbers, we 
shall have, by the preceding Case, 

Solid AG : P : : AE : AN. 

But, by hypothesis, we have 

Solid AG : solid AL : : AE : AO. 

Hence (Prop. IV., Cor., B. II.), 





136 


GEOMETRY. 


Solid AL : P : : AO : AN. 

But AO is greater than AN; hence the solid AL must be 
greater than P (Def. 2, B. II.); on the contrary, it is less, 
which is absurd. Therefore the solid AG can not be to the 
solid AL, as the line AE to a line greater than AI. 

In the same manner, it may be proved that the fourth term 
of the proportion can not be less than AI; hence it must be 
AI, and we have the proportion 

Solid AG : solid AL : : AE : AI. 

Therefore, right parallelopipeds, &c. 


PROPOSITION IX. THEOREM. 

Right parallelopipeds , having the same altitude , are to each 
other as their bases. 

Let AG, AN be two right parallelopipeds having the same 
altitude AE; then will they be to each other as their bases; 
that is, 

Solid AG : solid AN : : base ABCD : base AIKL. „ 

Place the two solids so that their 
surfaces may have the common 1 
angle BAE ; produce the plane 
LKNO till it meets the plane DCGH 
in the line PQ; a third parallel- 
opiped AQ will thus be formed, 
which may be compared with each 
of the parallelopipeds AG, AN. 1 
The two solids AG, AQ,, having 
the same base AEHD, are to each 
other as their altitudes AB, AL 
(Prop. VIII.); and the two solids B 

AQ, AN, having the same base ALOE, are to each other as 
their altitudes AD, AI. Hence we have the two proportions 
Solid AG : solid AQ : : AB : AL; 

Solid AQ : solid AN : : AD : AI. 

Hence (Prop. XI., Cor., B. II.), 

Solid AG : solid AN : : AB X AD : AL X AI. 

But ABxAD is the measure of the base ABCD (Prop. IV., 
Sch., B. IV.); and ALxAI is the measure of the base 
AIKL ; hence 

Solid AG : solid AN : ; base ABCD : base AIKL. 
Therefore, right parallelopipeds, &c. 


















BOOK VIII. 


137 


PROPOSITION X. THEOREM. 

Any two right parallelopipeds are to each other as the prod¬ 
ucts of their bases by their altitudes . 

Let AG, AQ be two right paral- M H 

Ielopipeds, of which the bases are 
the rectangles ABCD, AIKL, and 
the altitudes the perpendiculars AE, 

AP; then will the solid AG be to 
the solid AQ, as the product of 
ABCD by AE, is to the product of 
AIKL by AP. 

Place the two solids so that their 
surfaces may have the common an¬ 
gle BAE; produce the planes ne¬ 
cessary to form the third parallelo- 
piped AN, having the same base with AQ, and the same, alti¬ 
tude with AG. Then, by the last Proposition, we shall have 
Solid AG : solid AN : : ABCD : AIKL. 

But the two parallelopipeds AN, AQ, having the same base 
AIKL, are to each other as their altitudes AE, AP (Prop. 
VIII.); hence we have 

Solid AN : solid AQ : : AE : AP. 

Comparing these two proportions (Prop. XI., Cor., B. II.), 
we have 

Solid AG : solid AQ : : ABCDxAE : AIKLxAP. 

If instead of the base ABCD,we put its equal ABxAD, 
and instead of AIKL, we put its equal AIX AL, we shall have 

Solid AG : solid AQ : : AB X AD X AE : AI X AL X AP. 
Therefore, any two right parallelopipeds, &c. 

Scholium. Hence a right parallelopiped is measured by 
the product of its base and altitude, or the product of its three 
dimensions. 

It should be remembered, that by the product of two or 
more lines, we understand the product of the numbers which 
represent those lines; and these numbers depend upon the 
linear unit employed, which may be assumed at pleasure. 
If we take a foot as the unit of measure, then the number of 
feet in the length of the base, multiplied by the number of 
feet in its breadth, will give the number of square feet in the 
base. If we multiply this product by the number of feet in 
the altitude, it will give the number of cubic feet in the par¬ 
allelopiped. If we take an inch as the unit of measure, we 
shall obtain in the same manner the number of cubic inches 
in the parallelopiped. 














138 


GEOMETRY. 


PROPOSITION XI. THEOREM. 

The solidity of a prism is measured hy the product of its 
base by its altitude. 

For any parallelopiped is equivalent to a right parallelo- 
piped, having the same altitude and an equivalent base (Prop. 
VII.). But the solidity of the latter, is measured by the prod¬ 
uct of its base by its altitude; therefore the solidity of the 
former is also measured by the product of its base by its al¬ 
titude. 

Now a triangular prism is half of a~ parallelopiped having 
the same altitude and a double base (Prop. V.). But the 
solidity of the latter is measured by the product of its base by 
its altitude; hence a triangular prism is measured by the 
product of its base by its altitude. 

But any prism can be divided into as many triangular 
prisms of the same altitude, as there are triangles in the poly¬ 
gon which forms its base. Also, the solidity of each of these 
triangular prisms, is measured by the product of its base by 
its altitude; and since they all have the same altitude, the 
sum of these prisms will be measured by the sum of the tri¬ 
angles which form the bases, multiplied by the common alti¬ 
tude. Therefore, the solidity of any prism is measured by 
the product of its base by its altitude. 

Cor. If two prisms have the same altitude, the products of 
the bases by the altitudes,will be as the bases (Prop. VIII., 
B. II.) ; hence prisms of the same altitude are to each other as 
their bases. For the same reason, prisms of the same base 
are to each other as their altitudes ; and prisms generally are 
to each other as the products of their bases and altitudes. 


PROPOSITION XII. THEOREM. 

Similar prisms are to each other as the cubes of their homol¬ 
ogous edges. 

Let ABCDE-F, abcde-f be two similar prisms; then will 
the prism AD-F be to the prism ad-f, as AB 3 to ab% or as 
AF 3 to af. 

For the solids are to each other as the products of their 
bases and altitudes (Prop. XI., Cor.); that is, as ABODE X 
AF,to abcde X af. But since the prisms are similar, the bases 
are similar figures, and are to each other as the squares of 


BOOK VIII. 


139 



their homologous sides; that is, as AB 2 to ab\ Therefore, 
we have 

Solid FD : solid fd : : AB 2 x AF : ah' 2, X af. 

But since BF and bf are similar figures, their homologous 
sides are proportional; that is, 

AB : ab : : AF : af, 
whence (Prop. X., B. II.), 

AB 2 : «6 2 : : AF 2 : af, 
or (Prop. XI., B. II.), 

AB 2 x AF : ab'xaf : : AF 3 : af : : AB 3 : ab\ 

Hence (Prop. IV., B. II.), we have 

Solid FD : solid fd : : AB 3 : ab 3 : : AF 3 : af. 
Therefore, similar prisms, &c. 


PROPOSITION XIII. THEOREM. 


If a pyramid is cut by a plane parallel to its base, 

1st. The edges and the altitude will be dividedproportionally. 
2d. The section will be a polygon similar to the base. 


Let A-BCDEF be a pyramid cut by a 
plane bcdef parallel to its base, and let 
AH be its altitude; then will the sides 
AB, AC, AD, &c., with the altitude AH, 
be divided proportionally in b, c, d, e, f, 
h ; and the section bcdef will be similar to 
BCDEF. 

First. Since the planes FBC, fbc are 
parallel, their sections FB,fb with a third 
plane AFB are parallel (Prop. XII., B. 
VII.); therefore the triangles AFB, Afb 
are similar, and we have the proportion 
AF : Af: : AB : Ab. 
For the same reason, 

AB : Ab : : AC : Ac, 






























140 


GEOMETRY. 


and so for the other edges. Therefore the edges AB, AC, 
&c., are cut proportionally in b, c, &c. Also, since BH and 
bh are parallel, we have 

AH : Ah : : AB : A b. 

Secondly. Because fb is parallel to FB, be to BC, cd to CD, 
&c., the angle fbc is equal to FBC (Prop. XV., B. VII.), the 
angle bed is equal to BCD, and so on. Moreover, since the 
triangles AFB, A fb are similar, we have 
• FB :fb : : AB : Ab. 

And because the triangles ABC, A be are similar, we have 
AB : Ab : : BC : be. 

Therefore, by equality of ratios (Prop. IV., B. II.), 

FB :fb : : BC : be. 

For the same reason, 

BC : be : : CD : cd, and so on. 

Therefore the polygons BCDEF, bedef have their angles 
equal, each to each, and their homologous sides proportional; 
hence they are similar. Therefore, if a pyramid, &c. 

Cor. 1. If two pyramids, having the same altitude, and their 
bases situated in the same plane, are cut by a plane parallel to 
their bases, the sections will be to each other as the bases. 

Let A-BCDEF, A-MNO 
be two pyramids having 
the same altitude, and their 
bases situated in the same 
plane ; if these pyramids 
are cut by a plane parallel 
to the bases, the • sections 
bedef, mno will be to each 
other as the bases BCDEF, 

MNO. 

For, since the polygons 
BCDEF, bedef are similar, 
their surfaces are as the squares of the homologous sides BC, 
be (Prop. XXVI., B. IV.). But, by the preceding Proposition, 
BC : be : : AB : Ab. 

Therefore, BCDEF : bedef : : AB 2 : A6 2 . 

For the same reason, 

MNO : mno : : AM 2 : Am 2 . 

But since bedef and mno are in the same plane, we have 
AB : Ab : : AM : Am (Prop. XVI., B. VII.); 
consequently, BCDEF : bedef: : MNO : mno. 

Cor. 2. If the bases BCDEF, MNO are equivalent, the 
sections bedef mno will also be equivalent. 




BOOK VIII. 


141 


PROPOSITION XIV. THEOREM. 



The convex surface of a regular pyramid, is equal to the pe¬ 
rimeter of its base, multiplied by half the slant height. 

Let A-BDE be a regular pyramid, 
whose base is the polygon BCDEF, and 
its slant height AH ; then will its convex 
surface be equal to the perimeter BC+CD 
+DE, &c., multiplied by half of AH. 

The triangles AFB, ABC, ACD, &c., 
are all equal, for the sides FB, BC, CD, 

&c., are all equal (Def. 13); and since the 
oblique lines AF, AB, AC, &c., are all at 
equal distances from the perpendicular^ 
they are equal to each other (Prop. V., B.* 

VII.). Hence the altitudes of these sev- B C 
eral triangles are equal. But the area of the triangle AFB 
is equal to FB, multiplied by half of AH; and the same is 
true of the other triangles ABC, ACD, &c. Hence the sum 
of the triangles is equal to the sum of the bases FB, BC, CD, 
DE, EF, multiplied by half the common altitude AH; that 
is, the convex surface of the pyramid is equal to the pe¬ 
rimeter of its base, multiplied by half tl^e slant height. 

Cor. 1. The convex surface of a frustrum of a regular 
pyramid,is equal to the sum of the perimeters of its two bases, 
multiplied by half its slant height. 

Each side of a frustrum of a regular pyramid, as FB bf, is 
a trapezoid (Prop. XIII.). Now the area of this trapezoid is 
equal to the sum of its parallel sides FB, fb, multiplied by 
half its altitude H h (Prop. VII., B. IV.). But the altitude of 
each of these trapezoids is the same; therefore the area of 
all the trapezoids, or the convex surface of the frustrum, is 
equal to the sum of the perimeters of the two bases, multi¬ 
plied by half the slant height. 

Cor. 2. If the frustrum is cut by a plane parallel to the 
bases, and at equal distances from them, this plane must bi¬ 
sect the edges B6, Cc, &c. (Prop. XVI., B. IV.) ; and the area 
of each trapezoid is equal to its altitude, multiplied by the 
line which joins the middle points of its two inclined sides 
(Prop. VII., Cor., B. IV.). Hence the convex surface of a 
frustrum of a pyramid, is equal to its slant height, multiplied 
by the perimeter of a section at equal distances between the two 
bases. 



142 


GEOMETRY. 


PROPOSITION XV. THEOREM. 

Triangular pyramids, having equivalent bases and equal al¬ 
titudes ,are equivalent. 



Let A-BCD, a-bod be two triangular pyramids having 
equivalent bases BCD. bed , supposed to be situated in the 
same plane, and having the common altitude TB; then will 
the pyramid A-BCD be equivalent to the pyramid a-bed. 

For, if they are not equivalent, let the pyramid A-BCD 
exceed the pyramid a-bed by a prism whose base is BCD, 
and altitude BX. 

Divide the altitude BT into equal parts, each less than 
BX ; and through the several points of division,let planes be 
made to pass parallel to the base BCD, making the sections 
EFG, efg equivalent to each other (Prop. XIII., Cor. 2) ; 
also, HIK equivalent to hik, &c. 

From the point C, draw the straight line CR parallel to 
BE, meeting EF produced in R; and from D draw DS par¬ 
allel to BE, meeting EG in S. Join RS, and it is plain that 
the solid BCD-ERS is a prism lying partly without the pyr¬ 
amid. In the same manner, upon the triangles EFG, HIK, 
&c., taken as bases, construct exterior prisms, having for 
edges the parts EH, HL, &c., of the line AB. In like man¬ 
ner, on the bases efg, hik , hnn , &c., in the second pyramid, 
construct interior prisms, having for edges the corresponding 
parts of ab. It is plain that the sum of all the exterior prisms 














BOOK VIII. 


143 


of the pyramid A-BCD is greater than this pyramid; and, 
also, that the sum of all the interior prisms of the pyramid 
a-bcd is smaller than this pyramid. Hence the difference 
between the sum of all the exterior prisms, and the sum of 
all the interior ones, must be greater than the difference be¬ 
tween the two pyramids themselves. 

Now, beginning with the bases BCD, bed , the second ex¬ 
terior prism EFG-H is equivalent to the first interior prism 
efg-b , because their bases are equivalent, and they have the 
same altitude. For the same reason, the third exterior prism 
HIK-L and the second interior prism hik-e are equivalent; 
the fourth exterior and the third interior; and so on, to the 
last in each series. Hence all the exterior prisms of the pyr¬ 
amid A-BCD, excepting the first prism BCD-E, have equiv- 
lent corresponding ones in the interior prisms of the pyramid 
a-bcd. Therefore the prism BCD-E ip the difference be¬ 
tween the sum of all the exterior prisms of the pyramid 
A-BCD, and the sum of all the interior prisms of the pyr¬ 
amid a-bcd. But the difference between these two sets of 
prisms has been proved to be greater than that of the two 
pyramids ; hence the prism BCD-E is greater than the prism 
BCD-X ; which is impossible, for they have the same base 
BCD, and the altitude of the first, is less than BX, the altitude 
of the second. Hence the pyramids A-BCD, a-bcd are not 
unequal; that is, they are equivalent to each other. There¬ 
fore, triangular pyramids, &c. 


PROPOSITION XVI. THEOREM. 

Every triangular pyramid is the third part of a triangulai 
prism having the same base and the same altitude. 

Let E-ABC be a triangular pyramid, 
and ABC-DEF a triangular prism hav¬ 
ing the same base and the same altitude ; 
then will the pyramid be one third of the 
prism. 

Cut off from the prism the pyramid 
E-ABC by the plane EAC ; there will re¬ 
main the solid E-ACFD, which may be 
considered as a quadrangular pyramid 
whose vertex is E, and whose base is the 
parallelogram ACFD. Draw the diago¬ 
nal CD, and through the points C, D, E pass a plane, dividing 
the quadrangular pyramid into two triangular ones E-ACD, 
E-CFD. Then, because ACFD is a parallelogram, of which 



144 


GEOMETRY. 


CD is the diagonal, the triangle ACD is 
equal to the triangle CDF. Therefore 
the pyramid, whose base is the triangle 
ACD, and vertex the point E, is equiva¬ 
lent to the pyramid whose base is the tri¬ 
angle CDF, and vertex the point E. But 
the latter pyramid is equivalent to the 
pyramid E-ABC, for they have equal 
bases, viz., the triangles ABC, DEF, and 
the same altitude, viz., the altitude of the 
prism ABC-DEF. Therefore the three 
pyramids E-ABC, E-ACD, E-CDF, are equivalent to each 
other, and they compose the whole prism ABC-DEF ; hence 
the pyramid E-ABC is the third part of the prism which 
has the same base and the same altitude. 

Cor. The solidity of a triangular pyramid is measured by 
the product of its base by one third of its altitude. 



PROPOSITION XVlI. THEOREM. 


The solidity of every pyramid is measured hy the product of 
its base by one third of its altitude. 


Let A-BCDEF be any pyramid, whose 
base is the polygon BCDEF, and altitude 
AH; then will "the solidity of the pyramid 
be measured by BCDEF x ^AH. 

Divide the polygon BCDEF into triangles 
by the diagonals CF, DF; and let planes 
pass through these lines and the vertex A; 
they will divide the polygonal pyramid 
A-BCDEF into triangular pyramids, all 
having the same altitude AH. But each of 
these pyramids is measured by the product 
of its base by one third of its altitude (Prop. 

XVI., Cor.); hence the sum of the triangular pyramids, or 
the polygonal pyramid A-BCDEF, will be measured by the 
sum of the triangles BCF, CDF, DEF, or the polygon 
BCDEF, multiplied by one third of AH. Therefore every 
pyramid is measured by the product of its base by one third 
of its altitude. 

Cor. 1 . Every pyramid is one third of a prism having the 
same base and altitude. 

Cor. 2. Pyramids of the same altitude are to each other 
as their bases; pyramids of the same base are to each other 





BOOK VIII. 


145 


as their altitudes ; and pyramids generally are to each other 
as the products of their bases by their altitudes. 

Cor. 3. Similar pyramids are to each other as the cubes 
of their homologous edges. 

Scholium. The solidity of any polyedron may be found 
by dividing it into pyramids, by planes passing through one 
of its vertices. 


PROPOSITION XVIII. THEOREM. 

A frustrum of a 'pyramid is equivalent to the sum of three 
pyramids, having the same altitude as the frustrum, and whose 
bases are the lower base of the f rustrum, its upper base, and a 
mean proportional between them. 

Case first. When the base of the frustrum is a triangle. 

Let ABC-DEF be a frustrum of a tri- d 
angular pyramid. If a plane be made to 
pass through the points A, C, E, it will 
cut off the pyramid E-ABC, whose alti¬ 
tude is the altitude of the frustrum, and 
its base is ABC, the lower base of the 
frustrum. 

Pass another plane through the points A 
C, D, E ; it will cut off the pyramid 
C-DEF, whose altitude is that of the 
frustrum, and its base is DEF, the upper B 

base of the frustrum. 

To find the magnitude of the remaining pyramid E-ACD, 
draw EG parallel to AD ; join CG, DG. Then, because the 
two triangles AGC, DEF have the angles at A and D equal 
to each other, we have (Prop. XXIII., B. IV.) 

AGC : DEF : : AGxAC : DExDF, 

: : AC : DF, because AG is equal to DE. 
Also (Prop. VI., Cor. 1, B. IV.), 

ACB : ACG : : AB : AG or DE. 

But, because the triangles ABC, DEF are similar (Prop. 
XIII.), we have 

AB : DE : : AC : DF. 

Therefore (Prop. IV., B. II.), 

ACB : ACG : : ACG : DEF; 

that is, the triangle ACG is a mean proportional between 
ACB and DEF, the two bases of the frustrum. 

Now the pyramid E-ACD is equivalent to the pyramid 
G-ACD, because it has the same base and the same altitude; 
for EG is parallel to AD, and, consequently, parallel to the 




146 


GEOMETRY. 


plane ACD. But the pyramid G-ACD has the same altitude 
as the frustrum, and its base ACG is a mean proportional 
between the two bases of the frustrum. 

Case second. When the base of the frustrum is any polygon. 

Let BCDEF -bcdef be a 
frustrum of any pyramid. 

Let G-HIK be a trian¬ 
gular pyramid having the 
same altitude and an equiv¬ 
alent base with the pyramid 
A-BCDEF, and from it let 
a frustrum HlK-hik be cut 
off, having the same altitude 
with the frustrum BCDEF- C I 

bcdef. The entire pyramids are equivalent (Prop. XV.); 
and the small pyramids A -bcdef G-hik are also equivalent, 
for their altitudes are equal, and their bases are equivalent 
(Prop. XIII., Cor. 2.) Hence the two frustrums are equiva¬ 
lent, and they have the same altitude, with equivalent bases. 
But the frustrum HIK-Az/t has been proved to be equivalent 
to the sum of three pyramids, each having the same altitude 
as the frustrum, and whose bases are the lower base of the 
frustrum, its upper base, and a mean proportional between 
them. Hence the same must be true of a frustrum of any 
pyramid. Therefore, a frustrum of a pyramid, &c. 


A 


G 



PROPOSITION XIX. THEOREM. 

There can be but five regular polyedrons. 

Since the faces of a regular polyedron are regular poly¬ 
gons, they must consist of equilateral triangles, of squares, of 
regular pentagons, or polygons of a greater number of sides. 

First. If the faces are equilateral triangles, each solid an¬ 
gle of the polyedron may be contained by three of these tri¬ 



angles, forming the tetraedron; or by four, forming the oc- 
taedron; or by five, forming the icosaedron. 

No other regular polyedron can be formed with equilat¬ 
eral triangles; for six angles of these triangles amount to 




























BOOK vnr. 


147 


four right angles, and can not form a solid angle 
(Prop. XVIII., B. VII.). 

Secondly . If the faces are squares, their an¬ 
gles may be united three and three, forming 
the liexaedron , or cube. 

Four angles of squares amount to four right 
angles, and can not form a solid angle. 

Thirdly. If the faces are regular penta¬ 
gons, their angles may be united three and 
three, forming the regular dodecaedron. Four 
angles of a regular pentagon, are greater 
than four right angles, and can not form a 
solid angle. 

Fourthly. A regular polyedron can not be 
formed with regular hexagons, for three angles of a regular 
hexagon amount to four right angles. Three angles of a 
regular heptagon amount to more than four right angles; 
and the same is true of any polygon having a greater number 
of sides. 

Hence there can be but five regular polyedrons; three 
formed with equilateral triangles, one with squares, and one 
with pentagons. 








148 


GEOMETRY. 


BOOK IX. 


SPHERICAL GEOMETRY. 



Definitions. 

1. A sphere is a solid bounded by a curved surface, all the 
points of which are equally distant from a point within, call¬ 
ed the center. 

The sphere may be conceived to be de¬ 
scribed by the revolution of a semicircle 
ADB, about its diameter AB, which re¬ 
mains unmoved. 

2. The radius of a sphere, is a straight 
line drawn from the center to any point of 
the surface. The diameter , or axis , is a line 
passing through the center, and terminated 
each way by the surface. 

All the radii of a sphere are equal; all the diameters are 
also equal, and each double of the radius. 

3. It will be shown (Prop. I.), that every section of a 
sphere made by a plane is a circle. A great circle is a sec¬ 
tion made by a plane which passes through the center of the 
sphere. Any other section made by a plane is called a small 
circle. 

4. A plane touches a sphere,when it meets the sphere, but, 
being produced, does not cut it. 

5. The pole of a circle of a sphere, is a point in the surface 
equally distant from every point in the circumference of 
this circle. It will be shown (Prop. V.), that every circle, 
whether great or small, has two poles. 

6. A spherical triangle is a part of the sur¬ 
face of a sphere, bounded by three arcs of 
great circles, each of which is less than a semi¬ 
circumference. These arcs are called the sides 
of the triangle; and the angles which their 
planes make with each other, are the angles 
of the triangle. 

7. A spherical triangle is called right-angled , isosceles , or 
equilateral , in the same cases as a plane triangle. 




BOOK IX. 


149 


8. A spherical polygon is a part of the sur¬ 
face of a sphere bounded by several arcs of 
great circles. 



9. A lune is a part of the surface of a sphere in¬ 
cluded between the halves of two great circles. 

10. A spherical wedge , or ungula , is that portion 
of the sphere included between the same semicir¬ 
cles, and has the lune for its base. 


11. A spherical pyramid is a portion of the 
sphere included between the planes of a solid 
angle, whose vertex is at the center. The base 
of the pyramid is the spherical polygon inter¬ 
cepted by those planes. 


12. A zone is a part of the surface of a 
sphere included between two parallel planes. 

13. A spherical segment is a portion of the 
sphere included between two parallel planes. 

14. The bases of the segment are the sec¬ 
tions of the sphere; the altitude of the seg¬ 
ment, or zone, is the distance between the 
sections. One of the two planes may touch the sphere, in 
which case the segment has but one base. 

15. A spherical sector is a solid de- A 

scribed by the revolution of a circular 
sector, in the same manner as the 
sphere is described by the revolution 
of a semicircle. 

While the semicircle ADB, revolving 
round its diameter AB, describes a 
sphere, every circular sector, as ACE 
or ECD, describes a spherical sector. 

16. Two angles which are together 
equal to two right angles; or two arcs 



e^BB 





/ 

^ \ 

£ 


which are together equal to a semicircumference, are called 
the supplements of each other. 



















150 


GEOMETRY. 


PROPOSITION I. THEOREM. 

Every section of a sphere, made hy a plane , is a circle . 

Let ABD be a section, made by a 
plane, in a sphere whose center is C. 

From the point C draw CE perpendicu¬ 
lar to the plane ABD ; and draw lines 
CA, CB, CD, &c., to different points of 
the curve ABD which bounds the sec¬ 
tion. 

The oblique lines CA, CB, CD are 
equal, because they are radii of the 
sphere ; therefore they are equally distant from the perpen¬ 
dicular CE (Prop. V., Cor., B. VII.). Hence all the lines 
EA, EB, ED are equal; and, consequently, the section ABD 
is a circle, of which E is the center. Therefore, every sec¬ 
tion, &c. 

Cor. 1. If the section passes through the center of the 
sphere, its radius will be the radius of the sphere ; hence all 
great circles of a sphere are equal to each other. 

Cor. 2. Two great circles always bisect each other; for, 
since they have the same center, their common section is a 
diameter of both, and therefore bisects both. 

Cor. 3. Every great circle divides the sphere and its sur¬ 
face into two equal parts. For if the two hemispheres are 
separated and applied to each other, base to base, with their 
convexities turned the same way, the two surfaces must co¬ 
incide; otherwise there would be points in these surfaces un¬ 
equally distant from the center. 

Cor. 4. The center of a small circle, and that of the sphere, 
are in a straight line perpendicular to the plane of the small 
circle. 

Cor. 5. The circle which is furthest from the center is the 
least; for the greater the distance CE, the less is the chord 
AB, which is the diameter of the small circle ABD. 

Cor. 6. An arc of a great circle may be made to pass 
through any two points on the surface of a sphere ; for the 
two given points, together with the center of the sphere, 
make three points which are necessary to determine the posi¬ 
tion of a plane. If, however, the two given points were sit¬ 
uated at the extremities of a diameter, these two points and 
the center would then be in one straight line, and any num¬ 
ber of great circles might be made to pass through them. 



BOOK IX. 


151 


PROPOSITION II. THEOREM. 

Any two sides of a spherical triangle are greater than the 
third. 

Let ABC be a spherical triangle; any 
two sides as, AB, BC, are together greater 
than the third side AC. 

Let D be the center of the sphere ; and 
join AD, BD, CD. Conceive the planes 
ADB, BDC, CDA to be drawn, forming a 
solid angle at D. The angles ADB, BDC, 

CDA will be measured by AB, BC, CA, 
the sides of the spherical triangle. But 
when a solid angle is formed by three plane angles, the sum 
of any two of them is greater than the third (Prop. XVII., B. 

, VII.) ; hence any two of the arcs AB, BC, CA must be 
greater than the third. Therefore, any two sides, &c. 



PROPOSITION III. THEOREM. 

The shortest path from one point to another on the surface 
of a sphere , is the arc of a great circle joining the two given 
points . 

Let A and B be any two points on the surface of 
a sphere, and let ADB be the arc of a great circle 
which joins them; then will the line ADB be the 
shortest path from A to B on the surface of the 
sphere. 

For, if possible, let the shortest path from A to B 
pass through C, a point situated out of the arc of a 
great circle ADB. Draw AC, CB, arcs of great 
circles, and take BD equal to BC. 

By the preceding theorem, the arc ADB is less than AC+ 
CB. Subtracting the equal arcs BD and BC, there will re¬ 
main AD less than AC. Now the shortest path from B to C, 
whether it be an arc of a great circle, or some other line, is 
equal to the shortest path from B to D ; for, by revolving 
BC around B, the point C may be made to coincide with D, 
and thus the shortest path from B to C must coincide with 
the shortest path from B to D. But the shortest path from 
A to B was supposed to pass through C; hence the shortest 
path from A to C, can not be greater than the shortest path 
from A to D. 




152 


GEOMETRY. 


Now the arc AD has been proved to be less than AC; and 
therefore if AC be revolved about A until the point C falls 
on the arc ADB, the point C will fall between D and B. 
Hence the shortest path from C to A must be greater than 
the shortest path from D to A; but it has just been proved 
not to be greater, which is absurd. Consequently, no point 
of the shortest path from A to B, can be out of the arc of a 
great circle ADB. Therefore, the shortest path, &c. 


PROPOSITION IV. THEOREM. 

The sum of the sides of a spherical polygon , is less than the 
circumference of a great circle. 

Let ABCD be any spherical polygon; 
then will the sum of the sides AB, BC, CD, 

DA be less than the circumference of a 
great circle. 

Let E be the center of the sphere, and 
join AE, BE, CE, DE. The solid angle 
at E is contained by the plane angles AEB, 

BEC, CED, DEA, which together are less 
than four right angles (Prop. XVIII., B. 

VII.). Hence the sides AB, BC, CD, DA, 
which are the measures of these angles, are 
together less than four quadrants described with the radius 
AE; that is, than the circumference of a great circle. 
Therefore, the sum of the sides, &c. 



PROPOSITION V. THEOREM. 


The extremities of a diameter of a sphere , are the poles of all 
circles perpendicular to that diameter. 

Let AB be a diameter perpendicu¬ 
lar to CDE, a great circle of a sphere, 
and also to the small circle FGH; 
then will A and B, the extremities of 
the diameter, be the poles of both 
these circles. 

For, because AB is perpendicular 
to the plane CDE, it is perpendicular 
to every straight line Cl, DI, El, &c., 
drawn through its foot in the plane ; 
hence all the arcs AC, AD, AE, &c., are quarters of the cir- 







BOOK IX. 


153 


cumference. So, also, the arcs BC, BD, BE, &c., are quar¬ 
ters of the circumference; hence the points A and B are 
each equally distant from all the points of the circumference 
CDE ; they are, therefore, the poles of that circumference 
(Def. 5). 

Secondly. Because the radius AI is perpendicular to the 
plane of the circle FGH, it passes through Iv, the center of 
that circle (Prop. I., Cor. 4). Hence, if we draw the oblique 
lines AF, AG, AH, these lines will be equally distant from 
the perpendicular AK, and will be equal to each other (Prop. 
V., B. VII.). But since the chords AF, AG, AH are equal, 
the arcs are equal; hence the point A is a pole of the small 
circle FGH; and in the same manner it may be proved that 
B is the other pole. 

Cor. 1. The arc of a great circle AD, drawn from the pole 
to the circumference of another great circle CDE, is a qua¬ 
drant; and this quadrant is perpendicular to the arc CD. 
For, because AI is perpendicular to the plane CDI, every 
plane ADB which passes through the line AI is perpendicu¬ 
lar to the plane CDI (Prop. VI., B. VII.); therefore the. an¬ 
gle contained by these planes, or the angle ADC (Def. 6), is 
a right angle. 

Cor. 2. If it is required to find the pole of the arc CD, 
draw the indefinite arc DA perpendicular to CD, and take DA 
equal to a quadrant; the point A will be one of the poles of 
the arc CD. Or, at each of the extremities C and D, draw 
the arcs CA and DA perpendicular to CD ; the point of inter¬ 
section of these arcs will be the pole required. 

Cor. 3. Conversely, if the distance of the point A from 
each of the points C and D is equal to a quadrant, the point 
A will be the pole of the arc CD; and the angles ACD, 
ADC will be right angles. 

For, let I be the center of the sphere, and draw the radii 
AI, Cl, DI. Because the angles AIC, AID are right angles, 
the line AI is perpendicular to the two lines Cl, DI; it is, 
therefore, perpendicular to their plane (Prop. IV., B. VII.). 
Hence the point A is the pole of the arc CD (Prop. V.) ; and 
therefore the angles ACD, ADC are right angles (Cor. 1). 

Scholium. Circles may be drawn upon the surface of a 
sphere, with the same ease as upon a plane surface. Thus, 
by revolving the arc AF around the point A, the point F will 
describe the small circle FGH ; and if we revolve the qua¬ 
drant AC around the point A, the extremity C will describe 
the great circle CDE. 

If it is required to produce the arc CD, or if it is required 
to draw an arc of a great circle through the two points C 
and D, then from the points C and D as centers, with a radius 


154 


GEOMETRY. 


equal to a quadrant, describe two arcs intersecting each 
other in A. The point A will be the pole of the arc CD; 
and, therefore, if, from A as a center, with a radius equal to a 
quadrant, we'describe a circle CDE, it will be a great circle 
passing through C and D. 

If it is required to let fall a perpendicular from any point 
G upon the arc CD; produce CD to L, making DL equal to 
a quadrant; then from the pole L, with the radius DL, de¬ 
scribe the arc GD ; it will be perpendicular to CD. 


PROPOSITION VI. THEOREM. 


A plane,perpendicular to a diameter at its extremity, touches 
the sphere. 

Let ADB be a plane perpendicular A 
to the diameter DC at its extremity; 
then the plane ADB touches the 
sphere. 

Let E be any point in the plane 
ADB, and join DE, CE. Because CD 
is perpendicular to the plane ADB, it 
is perpendicular to the line AB (Def. 

1, B. VII.); hence the angle CDE is a right angle, and the 
line CE is greater than CD. Consequently, the point E lies 
without the sphere. Hence the plane ADB has only the point 
A in common with the sphere ; it therefore touches the sphere 
(Def. 4). Therefore, a-plane, &c. 

Cor. In the same manner, it may be proved that two 
spheres touch each other, when the distance between their 
centers is equal to the sum or difference of their radii; in 
which case, the centers and the point of contact lie in one 
straight line. 



PROPOSITION VII. THEOREM. 

The angle formed by two arcs of great circles, is equal to 
the angle formed by the tangents of those arcs at the point of 
their intersection ; and is measured by the arc of a great cir¬ 
cle described from its vertex as a pole, and included between its 
sides. 

Let BAD be an angle formed by two arcs of great circles; 
then will it be equal to the angle EAF formed by the tan- 




BOOK IX. 


155 



gents of these arcs at the point A, 
and it is measured by the arc DB de¬ 
scribed from the vertex A as a pole. 

For the tangent AE, drawn in the 
plane of the arc AB, is perpendicular 
to the radius*AC (Prop. IX., B. III.); 
also, the tangent AF, drawn in the 
plane of the arc AD, is perpendicular 
to the same radius AC. Hence the 
angle EAF is equal to the angle of the 
planes ACB, ACD (Def. 4, B. VII.), which is the same as 
that of the arcs AB, AD. 

Also, if the arcs AB, AD are each equal to a quadrant, the 
lines CB, CD will be perpendicular to AC, and the angle 
BCD will be equal to the angle of the planes ACB, ACD; 
hence the arc BD measures the angle of the planes, or the 
angle BAD. 

Cor. 1 . Angles of spherical triangles may be compared 
with each other by means of arcs of great circles described 
from their vertices as poles, and included between their 
sides ; and thus an angle can easily be made equal to a given 
angle. 

Cor. 2. If two arcs of great circles AC, 

DE cut each other, the vertical angles ABE, 

DBC are equal; for each is equal to the an¬ 
gle formed by the two planes ABC, DBE. 

Also, the two adjacent angles ABD, DBC 
are together equal to two right angles. 



PROPOSITION VIII. THEOREM. 

If from the vertices of a given spherical triangle , as poles , 
arcs of great circles are described , a second triangle is formed, 
whose vertices are poles of the sides of the given triangle. 

Let ABC be a spherical triangle; 
and from the points A, B, C, as poles, 
let great circles be described inter¬ 
secting each other in D, E, and F; 
then will the points D, E, and F be 
the poles of the sides of the triangle 
ABC. 

For, because the point A is the pole 
of the arc EF, the distance from A to 
E is a quadrant. Also, because the 
point C is the pole of the arc DE, the 








156 


GEOMETRY. 


distance from C to E is a quadrant. Hence the point E is at 
a quadrant’s distance from each of the points A and C ; it is, 
therefore, the pole of the arc AC (Prop. V., Cor. 3). In the 
same manner, it may be proved that D is the pole of the arc 
BC, and F the pole of the arc AB. 

Scholium. The triangle DEF is called the polar triangle 
of ABC; and so, also, ABC is the polar triangle of DEF. 

Several different triangles might be formed by producing 
the sides DE, EF, DF ; but we shall confine ourselves to the 
central triangle, of which the vertex D is on the same side 
of BC with the vertex A; E is on the same side of AC 
with the vertex B; and F is on the same side of AB with 
the vertex C. 


PROPOSITION IX. THEOREM. 

The sides of a spherical triangle , are the supplements of the 
arcs which measure the angles of its polar triangle ; and con¬ 
versely. 

Let ABC be a spherical triangle, 

DEF its polar triangle ; then will the 
side EF be the supplement of the arc 
which measures the angle A ; and 
the side BC is the supplement of the 
arc which measures the angle D. 

Produce the sides AB, AC, if ne¬ 
cessary, until they meet EF in G and 
H. Then, because the point A is the 
pole of the arc GH, the angle A is 
measured by the arc GH (Prop. VII.). 

Also, because E is the pole of the arc AH, the arc EH is a 
quadrant; and, because F is the pole of AG, the arc FG is a 
quadrant. Hence EH and GF, or EF and GH, are together 
equal to a semicircumference. Therefore EF is the supple¬ 
ment of GH, which measures the angle A. So, also, DF is 
the supplement of the arc which measures the angle B; and 
DE is the supplement of the arc which measures the angle C. 

Conversely. Because the point D is the pole of the arc BC, 
the angle D is measured by the arc IK. Also, because C is 
the pole of the arc DE, the arc IC is a quadrant; and, be¬ 
cause B is the pole of the arc DF, the arc BK is a quadrant. 
Hence IC and BK, or IK and BC, are together equal to a 
semicircumference. Therefore BC is the supplement of IK, 
which measures the angle D. So, also, AC is the supple¬ 
ment of the arc which measures the angle E ; and AB is the 
supplement of the arc which measures the angle F. 






BOOK IX. 


157 j 


PROPOSITION X. THEOREM. 

The sum of the angles of a spherical triangle,is greater than 
two , and less than six right angles. 

Let A, B, and C be the angles of a spherical triangle. 
The arcs which measure the angles A, B, and C, together 
with the three sides of the polar triangle, are equal to three 
semicircumferences (Prop. IX.). But the three sides of the 
polar triangle are less than two semicircumferences (Prop. 
IV.) ; hence the arcs which measure the angles A, B, and C 
are greater than one semicircumference; and, therefore, the 
angles A, B, and C are greater than two right angles. 

Also, because each angle of a spherical triangle is less than 
two right angles, the sum of the three angles must be less 
than six right angles. 

Cor. A spherical triangle may have two, or 
even three, right angles ; also two, or even 
three, obtuse angles. If a triangle have three 
right angles, each of its sides will be a qua¬ 
drant, and the triangle is called a quadrantal 
triangle. The quadrantal triangle is contain¬ 
ed eight times in the surface of the sphere. 



PROPOSITION XI. THEOREM. 


If two triangles on equal spheres are mutually equilateral , 
they are mutually equiangular. 


Let ABC, DEF be two triangles on equal spheres, having 
the sides AB equal to DE, AC to DF, and BC to EF; then 
will the angles also be equal, each to each. 




Let the centers of the spheres be G and H, and draw the 
radii GA, GB, GC, HD, HE, HF. A solid angle may be con¬ 
ceived as formed at G by the three plane angles AGB, AGC, 






158 


GEOMETRY. 


BGC; and another solid angle at H by the three plane an¬ 
gles DHE, DHF, EHF. Then, because the arcs AB, DE 
are equal, the angles AGB, DHE, which are measured by 
these arcs, are equal. For the same reason, the angles AGC, 
DHF are equal to each other; and, also, BGC equal to EHF. 



plane angles ; therefore the planes of these equal angles are 
equally inclined to each other (Prop. XIX., B. VII.). That 
is, the angles of the triangle ABC are equal to those of the 
triangle DEF, viz., the angle ABC to the angle DEF, BAC 
to EDF, and ACB to DFE. 

Scholium. It should be observed that the two triangles 
ABC, DEF do not admit of superposition, unless the three 
sides are similarly situated in both cases. Triangles which 
are mutually equilateral, but can not be applied to each other 
so as to coincide, are called symmetrical triangles. 


PROPOSITION XII. THEOREM. 

If two triangles on equal spheres are mutually equiangular , 
they are mutually equilateral. 

Denote by A and B two spherical triangles which are mu¬ 
tually equiangular, and by P and Q their polar triangles. 

Since the sides of P and Q, are the supplements of the arcs 
which measure the angles of A and B (Prop. IX.), P and 
Q must be mutually equilateral. Also, because P and Q are 
mutually equilateral, they must be mutually equiangular 
(Prop. XI.). But the sides of A and B are the supplements 
of the arcs which measure the angles of P and Q; and, 
therefore, A and B are mutually equilateral. 


PROPOSITION XIII. THEOREM. 

If two triangles on equal spheres have two sides , and the in¬ 
cluded angle of the one , equal to two sides and the included 
angle of the other , each to each , their third sides will he equal , 
and their other angles will he equal , each to each. 





BOOK IX. 


159 


Let ABC, DEF be two triangles, having the side AB equal 
to DE, AC equal to DF, and the angle BAC equal to the an¬ 
gle EDF; then will the side BC be equal to EF, the angle 
ABC to DEF, and ACB to DFE. 

If the equal sides in the two triangles are similarly sit¬ 
uated, the triangle ABC may be applied to the triangle DEF 
in the same manner as in plane tri- A 
angles (Prop. VI., B. I.); and the two v 
triangles will coincide throughout. Y\ 

Therefore all the parts of the one tri- \ \ 
angle, will be equal to the correspond- \ 
ing parts of the other triangle. / \ 

But if the equal sides in the two tri- / U F 
angles are not similarly situated, then j S' 
construct the triangle DF'E symmet- 
rical with DFE, having DF' equal to 
DF, and EF' equal to EF. The two triangles DEF', DEF, 
being mutually equilateral, are also mutually equiangular 
(Prop. XI.). Now the triangle ABC may be applied to the 
triangle DEF', so as to coincide throughout; and hence all 
the parts of the one triangle, will be equal to the correspond¬ 
ing parts of the other triangle. Therefore the side BC, be¬ 
ing equal to EF', is also equal to EF; the angle ABC, being 
equal to DEF', is also equal to DEF; and the angle ACB, 
being equal to DF'E, is also equal to DFE. Therefore, if 
two triangles, &c. 



PROPOSITION XIV. THEOREM. 


If two triangles on equal spheres have two angles, and the 
included side of the one, equal to two angles and the included side 
of the other, each to each, their third angles will he equal, and 
their other sides will he equal, each to each. 


If the two triangles ABC, DEF 
have the angle BAC equal to the an¬ 
gle EDF, the angle ABC equal to 
DEF, and the included side AB equal 
to DE; the triangle ABC can be 
placed upon the triangle DEF, or 
upon its symmetrical triangle DEF', 
so as to coincide. Hence the remain¬ 
ing parts of the triangle ABC, will be B 
equal to the remaining parts of the triangle DEF; that is, 
the side AC will be equal to DF, BC to EF, and the angle 
ACB to the angle DFE. Therefore, if two triangles, &c. 



160 


GEOMETRY. 


PROPOSITION XV. THEOREM. 



If two triangles on equal spheres are mutually equilateral, 
they are equivalent. 

Let ABC, DEF be two triangles 
which have the three sides of the 
one, equal to the three sides of the 
other, each to each, viz., AB to 
DE, AC to DF, and BC to EF; 
then will the triangle ABC be 
equivalent to the triangle DEF. 

Let G be the pole of the small 
circle passing through the three 
points A, B, C; draw the arcs GA, GB, GC ; these arcs will 
be equal to each other (Prop. V.). At the point E, make the 
angle DEH equal to the angle ABG; make the arc EH 
equal to the arc BG ; and join DH, FH. 

Because, in the triangles ABG, DEH, the sides DE, EH 
are equal to the sides AB, BG, and the included angle DEH 
is equal to ABG; the arc DH is equal to AG, and the angle 
DHE equal to AGB (Prop. XIII.). 

Now, because the triangles ABC, DEF are mutually equi¬ 
lateral, they are mutually equiangular (Prop. XI.); hence 
the angle ABC is equal to the angle DEF. Subtracting the 
equal angles ABG, DEH, the remainder GBC will be equal 
to the remainder HEF. Moreover, the sides BG, BC are 
equal to the sides EH, EF; hence the arc HF is equal to the 
arc GC, and the angle EHF to the angle BGC (Prop. XIII.). 

Now the triangle DEH may be applied to the triangle 
ABG so as to coincide. For, place DH upon its equal BG, 
and HE upon its equal AG, they will coincide, because the 
angle DHE is equal to the angle AGB; therefore the two 
triangles coincide throughout, and have equal surfaces. For 
the same reason, the surface HEF is equal to the surface 
GBC, and the surface DFH to the surface ACG. Hence 
ABG+GBC—ACG=DEH+EHF—DFH; 
or, ABC = DEF; 

that is, the two triangles ABC, DEF are equivalent. There¬ 
fore, if two triangles, &c. 

Scholium. The poles G and H might be situated within 
the triangles ABC, DEF; in which case it would be neces¬ 
sary to add the three triangles ABG, GBC, ACG to form the 
triangle ABC; and also to add the three triangles DEH, 






BOOK IX. 


161 

EHF, DFH to form the triangle DEF; otherwise the demon- 
stration would be the same as above. 

Cor. If two triangles on equal spheres, are mutually equi¬ 
angular, they are equivalent. They are also equivalent, if 
they have two sides, and the included angle of the one, equal 
to two sides and the included angle of the other, each to 
each ; or two angles and the included side of the one, equal 
to two angles and the included side of the other. 


PROPOSITION XVI. THEOREM. 


In an isosceles spherical triangle , the angles opposite the 
equal sides are equal; and , conversely , if two angles of a 
spherical triangle are equal, the triangle is isosceles. 

Let ABC be a spherical triangle, having 
the side AB equal to AC ; then will the angle 
ABC be equal to the angle ACB. 

From the point A draw the arc AD to the 
middle of the base BC. Then, in the two tri¬ 
angles ABD, ACD, the side AB is equal to 
AC, BD is equal to DC, and the side AD is -g 
common ; hence the angle ABD is equal to 
the angle ACD (Prop. XI.). 

Conversely. Let the angle B be equal to the 
angle C ; then will the side AC be equal to 
the side AB. 

For if the two sides are not equal to each 
other, let AB be the greater; take BD equal 
to AC, and join DC. Then, in the triangles 
DBC, ACB, the two sides BD, BC are equal to 
the two sides CA, CB, and the included angles 
DBC, ACB are equal; hence the angle DCB is equal to the 
angle ABC (Prop. XIII.). But, by hypothesis, the angle ABC 
is equal to ACB ; hence DCB is equal to ACB, which is ab¬ 
surd. Therefore AB is not greater than AC; and, in the 
same manner, it can be proved that it is not less ; it is, con¬ 
sequently, equal to AC. Therefore, in an isosceles spherical 
triangle, &c. 

Cor. The angle BAD is equal to the angle CAD, and the 
angle ADB to the angle ADC ; therefore each of the last two 
angles is a right angle. Hence the arc drawn from the vertex 
of an isosceles spherical triangle, to the middle of the base, is 
perpendicular to the base, and bisects the vertical angle. 




162 


GEOMETRY. 


PROPOSITION XVII. THEOREM. 

In a spherical triangle , the greater side is opposite the greater 
angle , and conversely. 

Let ABC be a spherical triangle, hav¬ 
ing the angle A greater than the angle 
B; then will the side BC be greater 
than the side AC. 

Draw the arc AD, making the angle 
BAD equal to B. Then, in the triangle 
ABD, we shall have AD equal to DB B 
(Prop. XVI.); that is, BC is equal to the sum of AD and DC. 
But AD and DC are together greater than AC (Prop. II.): 
hence BC is greater than AC. * 

Conversely. If the side BC is greater than AC, then will 
the angle A be greater than the angle B. For if the angle 
A is not greater than B, it must be either equal to it, or less. 
It is not equal; for then the side BC would be equal to AC 
(Prop. XVI.), which is contrary to the hypothesis. Neither 
can it be less; for then the side BC would be less than AC, 
by the first case, which is also contrary to the hypothesis. 
Hence the angle BAC is greater than the angle ABC. 
Therefore, in a spherical triangle, &c. 



PROPOSITION XVIII. THEOREM. 


The area of a lune is to the surface of the sphere , as the an¬ 
gle of the lune is to four right angles. 

Let ADBE be a lune, upon a sphere 
whose center is C, and the diameter 
AB; then will the area of the lune be 
to the surface of the sphere, as the an¬ 
gle DCE to four right angles, or as the 
arc DE to the circumference of a great 
'circle. 

First. When the ratio of the arc to 
the circumference can be expressed in 
whole numbers. 

Suppose the ratio of DE to DEFG to be as 4 to 25. Now 
if we divide the circumference DEFG in 25 equal parts, DE 
will contain 4 of those parts. If we join the pole A and the 
several points of division, by arcs of great circles, there will 









BOOK IX. 


163 


be formed on the hemisphere ADEFG, 25 triangles, all equal 
to each other, being mutually equilateral. The entire sphere 
will contain 50 of these small triangles, and the lune ADBE 
8 of them. Hence the area of the lune is to the surface of 
the sphere, as 8 to 50, or as 4 to 25; that is, as the arc DE 
to the circumference. 

Secondly. When the ratio of the arc to the circumference, 
can not be expressed in whole numbers, it may be proved, as 
in Prop. XIV., B. III., that the lune is still to the surface of 
the sphere, as the angle of the lune to four right angles. 

Cor. 1 . On equal spheres, two lunes are to each other as 
the angles included between their planes. 

Cor. 2. We have seen that the entire surface of the sphere 
is equal to eight quadrantal triangles (Prop. X., Cor.). If 
the area of the quadrantal triangle be represented by T, the 
surface of the sphere will be represented by 8T. Also, if we 
take the right angle for unity, and represent the angle of the 
lune by A, we shall have the proportion 

4 : A : : 8T : area of the lune. 

8 A X T 

Hence the area of the lune is equal to —-—, or 2A X T. 


Cor. 3. The spherical ungula, comprehended by the planes 
ADB, AEB, is to the entire sphere, as the angle DCE is to 
four right angles. For the lunes being equal, the spherical 
ungulas will also be equal; hence, in equal spheres, two un- 
gulas are to each other as the angles included between their 
planes. 


PROPOSITION XIX. THEOREM. 

If two great circles intersect each other on the surface of a 
hemisphere , the sum of the opposite triangles thus formed, is 
equivalent to a lune , whose angle is equal to the inclination of 
the two circles. 

Let the great circles ABC, DBC in 
tersect each other on the surface of 
the hemisphere BADCE ; then will the 
sum of the opposite triangles ABD, 

CBE be equivalent to a lune whose a 
angle is CBE. 

For, produce the arcs BC, BE till 
they meet in F; then will BCF be a 
semicircumference, as also ABC. Sub¬ 
tracting BC from each, we shall have CF equal to AB. For 
the same reason EF is equal to DB, and CE is equal to AD. 





164 


GEOMETRY. 


Hence the two triangles ABD, CFE are mutually equilat¬ 
eral ; they are, therefore, equivalent (Prop. XV.). But the 
two triangles CBE, CFE compose the lune BCFE, whose an¬ 
gle is CBE; hence the sum of the triangles ABD, CBE is 
equivalent to the lune whose angle is CBE. Therefore, if 
two great circles, &c. 


PROPOSITION XX. THEOREM. 

The surface of a spherical triangle is measured by the ex¬ 
cess of the sum of its angles above two right angles , multiplied 
by the quadrantal triangle . 

Let ABC be any spherical triangle; its 
surface is measured by the sum of its an¬ 
gles A, B, C diminished by two right an¬ 
gles, and multiplied by the quadrantal tri¬ 
angle. 

Produce the sides of the triangle ABC, 
until they meet the great circle DEG, 
drawn without the triangle. The two 
triangles ADE, AGH are together equal 
to the lune whose angle is A (Prop. XIX.) ; and this lune is 
measured by 2A X T (Prop. XVIII., Cor. 2). Hence we have 
ADE+AGH=2A XT. 

For the same reason, 

BFG+BDI =2BxT; 
also, CHI+CEF=2 CxT. 

But the sum of these six triangles exceeds the surface of 
the hemisphere, by twice the triangle ABC; and the hemi¬ 
sphere is represented by 4T ; hence we have 

4T+2ABC = 2A X T+2B X T+2C X T ; 
or, dividing by 2, and then subtracting 2T from each of 
these equals, we have 

ABC = A x T-f B X T+C xT—2T, 

= (A+B+C—2) XT. 

Hence every spherical triangle is measured by the sum of 
its angles diminished by two right angles, and multiplied by 
the quadrantal triangle. 

Cor. If the sum of the three angles of a triangle is equal 
to three right angles, its surface will be equal to the quad¬ 
rantal triangle; if the sum is equal to four right angles, the 
surface of the triangle will be equal to two quadrantal trian¬ 
gles ; if the sum is equal to five right angles, the surface will 
be equal to three quadrantal triangles, etc. 




BOOK IX. 


165 


PROPOSITION XXI. THEOREM. 

The surface of a spherical polygon is measured by the sum 
of its angles , diminished by as many times two right angles as 
it has sides less two , multiplied by the quadrantal triangle. 

Let ABODE be any spherical polygon. 

From the vertex B draw the arcs BD, 

BE to the opposite angles; the polygon 
will be divided into as many triangles as 
it has sides, minus two. But the surface 
of each triangle is measured by the sum 
of its angles minus two right angles, mul¬ 
tiplied by the quadrantal triangle. Also, 
the sum of all the angles of the triangles, is equal to the sum 
of all the angles of the polygon; hence the surface of the 
polygon is measured by the sum of its angles, diminished by 
as many times two right angles as it has sides less two, mul¬ 
tiplied by the quadrantal triangle. 

Cor. If the polygon has five sides, and the sum of its an¬ 
gles is equal to seven right angles, its surface will be equal to 
the quadrantal triangle; if the sum is equal to eight right an¬ 
gles, its surface will be equal to two quadrantal triangles ; if 
the sum is equal to nine right angles, the surface will be 
equal to three quadrantal triangles, etc. 



166 


GEOMETRY. 


BOOK X. 

THE THREE ROUND BODIES. 

Definitions. 

1. A cylinder is a solid described by the revolu¬ 
tion of a rectangle about one of its sides, which 
remains fixed. The bases of the cylinder are the 
circles described by the two revolving opposite 
sides of the rectangle. 

2. The axis of a cylinder is the fixed straight 
line about which the rectangle revolves. The op¬ 
posite side of the rectangle describes the convex 
surface. 

3. A cone is a solid described by the revolu¬ 
tion of a right-angled triangle about one of the 
sides containing the right angle, which side re¬ 
mains fixed. The base of the cone is the cir¬ 
cle described by that side containing the right 
angle, which.revolves. 

4. The axis of a cone is the fixed straight 
line about which the triangle revolves. The 
hypothenuse of the triangle describes the convex surface. 
The side of the cone is the distance from the vertex to the 
circumference of the base. 

5. A frustrum of a cone is the part of a cone next the 
base, cut off by a plane parallel to the base. 

6. Similar cones and cylinders are those which have their 
axes and the diameters of their bases proportionals. 




PROPOSITION I. THEOREM. 

The convex surface of a cylinder is equal to the product of 
its altitude by the circumference of its base. 

Let ACE-G be a cylinder whose base is the circle ACE 
and altitude AG; then will its convex surface be equal to 
the product of AG by the circumference ACE. 




















BOOK X. 


167 



In the circle ACE inscribe the regular 
polygon ABCDEF; and upon this polygon 
let a right prism be constructed of the same 
altitude with the cylinder. The edges AG, 

BH, CK, &c., of the prism, being perpen¬ 
dicular to the plane of the base, will be con¬ 
tained in the convex surface of the cylinder. 

The convex surface of this prism is equal to 
the product of its altitude by the perimeter 
of its base (Prop. I., B. VIII.). Let, now, 
the arcs which subtend the sides AB, BC, &c., be bisected, 
and the number of sides of the polygon be indefinitely in¬ 
creased ; its perimeter will approach the circumference of the 
circle, and will be ultimately equal to it (Prop. XI., B. VI.); 
and the convex surface of the prism will become equal to 
the convex surface of the cylinder. But whatever be the 
number of sides of the prism, its convex surface is equal to 
the product of its altitude by the perimeter of its base ; hence 
the convex surface of the cylinder is equal to the product of 
its altitude by the circumference of its base. 

Cot\ If A represent the altitude of a cylinder, and R the 
radius of its base, the circumference of the base will be repre¬ 
sented by 2nR (Prop. XIII., Cor. 2, B. VI.); and the convex 
surface of the cylinder by 27rRA. 


PROPOSITION II. THEOREM. 

The solidity of a cylinder' is equal to the product of its base 
by its altitude . 

Let ACE-G be a cylinder whose base is 
the circle ACE and altitude AG ; its solidity 
is equal to the product of its base by its al¬ 
titude. 

In the circle ACE inscribe the regular 
polygon ABCDEF ; and upon this polygon 
let a right prism be constructed of the same 
altitude with the cylinder. The solidity of 
this prism is equal to the product of its base 
by its altitude (Prop. XI., B. VIII.). Let, 
now, the number of sides of the polygon be indefinitely in¬ 
creased ; its area will become equal to that of the circle, and 
the solidity of the prism becomes equal to that of the cylinder*. 
But whatever be the number of sides of the prism, its solidity 
is equal to the product of its base by its altitude; hence the 
solidity of a cylinder is equal to the product of its base by its 
altitude. 



















168 


GEOMETRY. 


Cor. 1 . If A represent the altitude of a cylinder, and R 
the radius of its base, the area of the base will be represent¬ 
ed by ttR 2 (Prop. XIII., Cor. 3, B. VI.); and the solidity of 
the cylinder will be 7tR 3 A. 

Cor. 2. Cylinders of the same altitude, are to each other 
as their bases; and cylinders of - the same base, are to each 
other as their altitudes. 

Cor. 3. Similar cylinders are to each other as the cubes 
of their altitudes, or as the cubes of the diameters of their 
bases. For the bases are as the squares of their diameters; 
and since the cylinders are similar, the diameters of the bases 
are as their altitudes (Def. 6). Therefore the bases are as 
the squares of the altitudes; and hence the products of the 
bases by the altitudes, or the cylinders' themselves, will be as 
the cubes of the altitudes. 


PROPOSITION III. THEOREM. 



The convex surface of a cone is equal to the product of half 
its side , by the circumference of its base. 

Let A-BCDEFG be a cone whose base is 
the circle BDEG, and its side AB ; then will 
its convex surface be equal to the product 
of half its side by the circumference of the 
circle BDF. 

In the circle BDF inscribe the regular 
polygon BCDEFG; and upon this polygon 
let a regular pyramid be constructed having 
A for its vertex. The edges of this pyramid 
will lie in the convex surface of the cone. 

From A draw AH perpendicular to CD, one of the sides of 
the polygon. The convex surface of the pyramid is equal to 
the product of half the slant height AH by the perimeter of 
its base (Prop. XIV., B. VIII.). Let, now, the arcs which 
subtend the sides BC, CD, &c., be bisected, and the number 
of sides of the polygon be indefinitely increased, its perimeter 
will become equal to the circumference of the circle, the slant 
height AH becomes equal to the side of the cone AB, and 
the convex surface of the pyramid becomes equal to the con¬ 
vex surface of the cone. But, whatever be the number of 
faces of the pyramid, its convex surface is equal to the prod¬ 
uct of half its slant height by the perimeter of its base ; hence 
the convex surface of the cone, is equal to the product of 
half its side by the circumference of its base. 

Cor. If S represent the side of a cone, and R the radius 


BOOK X. 


169 


of its base, then the circumference of the base will be repre¬ 
sented by 2 ttR, and the convex surface of the cone bv 
2 ttRxiS, or ttRS. 


PROPOSITION IV. THEOREM. 


The convex surface of a frustrum of a cone, is equal to the 
product of its side, by half the sum of the circumferences of its 
two bases. 

Let BDF-&J/' be a frustrum of a cone 
whose bases are BDF, bdf and B b its 
side; its convex surface is equal to the 
product of B b by half the sum of the cir¬ 
cumferences BDF, bdf 

Complete the cone A-BDF to which the 
frustrum belongs, and in the circle BDF 
inscribe the regular polygon BCDEFG; 
and upon this polygon let a regular pyra¬ 
mid be constructed having A for its ver¬ 
tex. Then will be BDF-fo/f be a frus¬ 
trum of a regular pyramid, whose convex 
surface is equal to the product of its slant height by half the 
sum of the perimeters of its two bases (Prop. XIV., Cor. 1, B. 
VIII.). Let, now, the number of sides of the polygon be in¬ 
definitely increased, its perimeter will become equal to the 
circumference of the circle, and the convex surface of the 
pyramid will become equal to the convex surface of the cone. 
But, whatever be the number of faces of the pyramid, the 
convex surface of its frustrum is equal to the product of its 
slant height, by half the sum of the perimeters of its two bases. 
Hence the convex surface of a frustrum of a cone, is equal to 
the product of its side by half the sum of the circumferences 
of its two bases. 

Cor. It was proved (Prop. XIV., Cor. 2, B. VIII.), that the 
convex surface of a frustrum of a pyramid, is equal to the 
product of its slant height, by the perimeter of a section at 
equal distances between its two bases ; hence the convex sur¬ 
face of a frustrum of a cone is equal to the product of its side , 
by the circumference of a section at equal distances between the 
two bases. 








170 


GEOMETRY. 


PROPOSITION- V. THEOREM. 

The solidity of a cone is equal to one third of the product of 
its base and altitude. 

Let A-BCDF be a cone whose base is the 
circle BCDEFG, and AH its altitude ; the 
solidity of the cone will be equal to one third 
of the product of the base BCDF by the al¬ 
titude AH. 

In the circle BDF inscribe a regular poly¬ 
gon BCDEFG, and construct a pyramid 
whose base is the polygon BDF, and having b£' v / 
its vertex in A. The solidity of this pyra- 
mid is equal to one third of the product of 
the polygon BCDEFG by its altitude AH (Prop. XVII., B. 
VIII.). Let, now, the number of sides of the polygon be in¬ 
definitely increased; its area will become equal to the area 
of the circle, and the solidity of the pyramid will become 
equal to the solidity of the cone. But, whatever be the 
number of faces of the pyramid, its solidity is equal to one 
third of the product of its base and altitude; hence the solidity 
of the cone is equal to one third of the product of its base and 
altitude. 

Cor. 1 . Since a cone is one third of a cylinder having the 
same base and altitude, it follows that cones of equal alti¬ 
tudes are to each other as their bases; cones of equal bases 
are to each other as their altitudes; and similar cones are as 
the cubes of their altitudes, or as the cubes of the diameters 
of their bases. 

Cor. 2. If A represent the altitude of a cone, and R the 
radius of its base, the solidity of the cone will be represented 
by 7rR 2 x JA, or ^R 2 A. 



PROPOSITION VI. THEOREM. 

A frustrum of a cone is equivalent to the sum of three cones, 
having the same altitude with the frustrum , and whose bases 
are the lower base of the frustrum , its upper base , and a mean 
proportional between them. 

Let BT>F-bdf be any frustrum of a cone. Complete the 
cone to which the frustrum belongs, and in the circle BDF 
inscribe the regular polygon BCDEFG; and upon this poly- 



BOOK X. 


171 


gon let a regular pyramid be constructed 
having its vertex in A. Then will 
BCDEFG-frccfc/g be a frustrum of a reg¬ 
ular pyramid, whose solidity is equal to 
three pyramids having the same altitude 
with the frustrum, and whose bases are 
the lower base of the frustrum, its upper 
base, and a mean proportional between 
them (Prop. XVIII., B. VIII.). Let, now, 
the number of sides of the polygon be in¬ 
definitely increased, its area will become 
equal to the area of the circle, and the 
frustrum of the pyramid will become -the frustrum of a cone. 
Hence the frustrum of a cone is equivalent to the sum of three 
cones, having the same altitude with the frustrum, and whose 
bases are the lower base of the frustrum, its upper base, and 
a mean proportional between them. 


A 



PROPOSITION VII. THEOREM. 


The surface of a sphere is equal to the product of its diame¬ 
ter hy the circumference of a great circle. 

Let ABDF be the semicircle by the revo¬ 
lution of which the sphere is described. In¬ 
scribe in the semicircle a regular semi-poly¬ 
gon ABCDEF, and from the points B, C, D, 

E let fall the perpendiculars BG, CH, DK, 

EL upon the diameter AF. If, now, the 
polygon be revolved about AF, the lines AB, 

EF will describe the convex surface of two d' 
cones; and BC, CD, DE will describe the 
convex surface of frustrums of cones. 

From the center I, draw IM perpendicular 
to BC ; also, draw MN perpendicular to AF, 
and BO perpendicular to CH. Let circ. MN represent the 
circumference of the circle described by the revolution of 
MN. Then the surface described by the revolution of BC, 
will be equal to BC, multiplied by circ. MN (Prop. IV., 
Cor.). 

Now, the triangles IMN, BCO are similar, since their sides 
are perpendicular to each other (Prop. XXI., B. IV.); whence 
BC : BO or GH : : IM : MN, 

: : circ. IM : circ. MN. 

Hence (Prop. I., B. II.). 

BC Xcirc. MN=GH Xc?Vc. IM. 












172 


GEOMETRY. 



Therefore the surface described by BC, is 
equal to the altitude GH, multiplied by circ. 

IM, or the circumference of the inscribed 
circle. 

In like manner, it may be proved that the 
surface described by CD is equal to- the alti¬ 
tude HK, multiplied by the circumference of 
the inscribed circle; and the same may be 
proved of the other sides. Hence the entire 
surface described by ABCDEF is equal to 
the circumference of the inscribed circle, mul¬ 
tiplied by the sum of the altitudes AG, GH, 

HK, KL, and LF; that, is, the axis of the polygon. 

Let, now, the arcs AB, BC, &c., be bisected, and the num¬ 
ber of sides of the polygon be indefinitely increased, its pe¬ 
rimeter will coincide with the circumference of the semicircle, 
and the perpendicular IM will become equal to the radius of 
the sphere; that is, the circumference of the inscribed circle 
will become the circumference of a great circle. Hence the 
surface of a sphere is equal to the product of its diameter by 
the circumference of a great circle. 

Cor. 1 . The area of a zone is equal to the product of its al¬ 
titude by the circumference of a great circle. 

For the surface described by the lines BC, CD is equal to 
the altitude GK, multiplied by the circumference of the in¬ 
scribed circle. But when the number of sides of the polygon 
is indefinitely increased, the perimeter BC-f-CD becomes the 
are BCD, and the inscribed circle becomes a great circle. 
Hence the area of the zone produced by the revolution of 
BCD, is equal to the product of its altitude GK by the cir¬ 
cumference of a great circle. 

Cor. 2. The area of a great circle is equal to the prod¬ 
uct of its circumference by half the radius (Prop. XII., B. 
VI.), or one fourth of the diameter; hence the surface of a 
sphere is equivalent to four of its great circles. 

Cor. 3. The surface of a sphere is equal to the convex sur¬ 
face of the circumscribed cylinder. 

For the latter is equal to the product of its 
altitude by the circumference of its base. But 
its base is equal to a great circle of the sphere, 
and its altitude to the diameter; hence the 
convex surface of the cylinder, is equal to the 
product of its diameter by the circumference 
of a great circle, which is also the measure of 
the surface of a sphere. 

Cor. 4. Two zones upon equal spheres, are to each other 
as their altitudes; and any zone is to the surface of its 














BOOK X. 


173 


sphere, as the altitude of the zone is to the diameter of the 
sphere. 

Cor. 5. Let R denote the radius of a sphere, D its diame¬ 
ter, C the circumference of a great circle, and S the surface 
of the sphere, then we shall have 

C = 2rrR, or nD (Prop. XIII., Cor. 2, B. VI.). 

Also, S=2 ttR X 2R = 4rrR a , or nD\ 

If A represents the altitude of a zone, its area will be 
27rRA. 


PROPOSITION VIII. THEOREM. 

The solidity of a sphere is equal to one third the product of 
its surface by the radius. 

Let ACEG be the semicircle by the revo¬ 
lution of which the sphere is described. In¬ 
scribe in the semicircle a regular semi-poly¬ 
gon ABCDEFG, and draw the radii BO, 

CO, DO, &c. 

The solid described by the revolution of 
the polygon ABCDEFG about AG, is com¬ 
posed of the solids formed by the revolu¬ 
tion of the triangles ABO, BCO, CDO, &c., 
about AG. 

First. To find the value of the solid form¬ 
ed by the revolution of the triangle ABO. 

From O draw OH perpendicular to AB, 
and from B draw BK perpendicular to AO. 

The two triangles ABK, BKO, in their revolution about AO, 
will describe two cones having a common base, viz., the cir¬ 
cle whose radius is BK. Let area BK represent the area 
of the circle described by the revolution of BK. Then the 
solid described by the triangle ABO will be represented by 
Area BKxjAO (Prop. V.). 

Now the convex surface of a cone is expressed by ttRS 
(Prop. III., Cor.); and the base of the cone by ttR\ Hence 
the convex surface : base : : 7 tRS : ttR 2 , 

: : S : R (Prop. VIII., B. II.). 

But AB describes the convex surface of a cone, of which 
BK describes the base ; hence 
the surface described by AB : area BK : : AB : BK 

: : AO : OH, 

because the triangles ABK, AHO are similar. Hence 
Area BKx AO= OHx surface described by AB, 
or Area BK x |AO= JOH x surface described by AB. 





174 


GEOMETRY. 


But we have proved that the solid de¬ 
scribed by the triangle ABO, is equal to 
area BKx^AO; it is, therefore, equal to 
iOH x surface described by AB. 

Secondly. To find the value of the solid 
formed by the revolution of the triangle 
BCO. 





( \ \ 

c/ 


JZ 

Tit. 


Up . 

u 


E \ 



in L. It is evident, from the preceding 
demonstration, that the solid described by 
the triangle LCO is equal to 

iOM x surface described by LC ; 
and the solid described by the triangle LBO 
is equal to 

iOM X surface described by LB ; 
hence the solid described by the triangle BCO is equal to 
iOM X surface described by BC. 

In the same manner, it may be proved that the solid de¬ 
scribed by the triangle CDO is equal to 

ION x surface described by CD ; 
and so on for the other triangles. But the perpendiculars 
OH, OM, ON, &c., are all equal; hence the solid described 
by the polygon ABCDEFG,is equal to the surface described 
by the perimeter of the polygon, multiplied by |OH. 

Let, now, the number of sides of the polygon be indefinite¬ 
ly increased, the perpendicular OH will become the radius 
OA, the perimeter ACEG will become the semi-circumference 
ADG, and the solid described by the polygon becomes a 
sphere; hence the solidity of a sphere is equal to one third 
of the product of its surface by the radius. 

Cor. 1. The solidity of a spherical sector is equal to the prod¬ 
uct of the zone which forms its base , by one third of its radius. 

For the solid described by the revolution of BCDO is 
equal to the surface described by BC+CD, multiplied by 
lOM. But when the number of sides of the polygon is in¬ 
definitely increased, the perpendicular OM becomes the 
radius OB, the quadrilateral BCDO becomes the sector 
BDO, and the solid described by the revolution of BCDO 
becomes a spherical sector. Hence the solidity of a spheri¬ 
cal sector is equal to the product of the zone which forms its 
base, by one third of its radius. 

Cor. 2. Let R represent the radius of a sphere, D its di¬ 
ameter, S its surface, and V its solidity, then we shall have 
S=47 tR 3 or 7rD 2 (Prop. VII., Cor. 5). 

Also, V=|RxS = IttR 3 or i7rD 3 ; 

hence the solidities of spheres are to each other as the cubes of 
their radii. 






BOOK X. 


17f> 

If we put A to represent the altitude of the zone which 
forms the base of a sector, then the solidity of the sector will 
be represented by 

2ttR A x ~R = fuR 2 A. 

Cor. 3. Every sphere is two thirds of the circumscribed 
cylinder. 

For, since the base of the circumscribed cylinder is equal 
to a great circle, and its altitude to a diameter, the solidity 
of the cylinder is equal to a great circle, multiplied by the 
diameter (Prop. II.). But the solidity of a sphere is equal 
to four great circles, multiplied by one third of the radius ; or 
one great circle, multiplied by £ of the radius, or f of the 
diameter. Hence a sphere is two thirds of the circumscribed 
cylinder. 


PROPOSITION IX. THEOREM. 


A spherical segment with one base, is equivalent to half of 
a cylinder having the same base and altitude, plus a sphere 
whose diameter is the altitude of the segment. 



Let BD be the radius of the base of the 
segment, AD its altitude, and let the segment 
be generated by the revolution of the circu¬ 
lar half segment AEBD about the axis AC. 

Join CB, and from the center C draw CF per¬ 
pendicular to AB. 

The solid generated by the revolution of 
the segment AEB, is equal to the difference of the solids gen¬ 
erated by the sector ACBE, and the triangle ACB. Now, 
the solid generated by the sector ACBE is equal to 
|7 tCB 2 xAD (Prop. VIII., Cor. 2). 

And the solid generated by the triangle ACB, by Prop. VIII., 
is equal to £CF, multiplied by th.e convex surface described 
by AB, which is 2rrCF X AD (Prop. VII.), making for the solid 
generated by the triangle ACB, 

IttCPxAD. 

Therefore the solid generated by the segment AEB, is equal 
to |7 tADx(CB 2 -CF 2 ), 

or |7 tAD x BF 2 ; 

that is, ^7 tAD x AB 2 , 

because CB 2 —CF 2 is equal to BF 2 , and BF 2 is equal to one 


fourth of AB 2 . 

Now the cone generated by the triangle ABD is equal to 
^7 tADxBD 2 (Prop. V., Cor. 2). 

Therefore the spherical segment in question, which is the 
sum of the solids described by AEB and ABD, is equal to 




176 


GEOMETRY. 


i7rAD(2BD 2 +AB 2 ); 
that is, |rrAD(3BD 2 -j- AD 2 ), 

because AB 2 is equal to BD 2 + AD 2 . 

This expression may be separated into the two parts 
^AD x BD 2 , 

and irrAD 3 . 

The first part represents the solid¬ 
ity of a cylinder having the same 
base with the segment and half its 
altitude (Prop. II.); the other part 
represents a sphere, of which AD is 
the diameter (Prop. VIII., Cor. 2). 
segment, &c. 

Cor. The solidity of the spherical seg¬ 
ment of two bases, generated by the revolu¬ 
tion of BCDE about the axis AD, may be 
found by subtracting that of the segment of 
one base generated by ABE, from that of the 
segment of one base generated by ACD. 



Therefore, a spherical 


....A 


r 








CONIC SECTIONS. 


There are three curves whose properties are extensively 
applied in Astronomy, and many other branches of science, 
which, being the sections of a cone made by a plane in dif¬ 
ferent positions, are called the conic sections. These are 
The Parabola , 

The Ellipse , and 
The Hyperbola. 


PARABOLA. 

Definitions. 

1. A parabola is a plane curve, every point of which is 
equally distant from a fixed point, and a given straight line. 

2. The fixed point is called the focus of the parabola, and 
the given straight line is called the directrix. 

Thus, if F be a fixed point, and BC a 
given line, and the point A move about*F 
in such a manner, that its distance from F 
is always equal to the perpendicular dis¬ 
tance from BC, the point A will describe 
a parabola, of which F is the focus, and 
BC the directrix. 

3. A diameter is a straight line drawn 
through any point of the curve perpen¬ 
dicular to the directrix. The vertex of 
the diameter is the point in which it cuts 
the curve. 

Thus, through any point of the curve, as A, draw a line 
DE perpendicular to the directrix BC; DE is a diameter of 
the parabola, and the point A is the vertex of this diameter. 

4. The axis of the parabola is the diameter which passes 
through the focus; and the point in which it cuts the curve 
is called the principal vertex. 

Thus, draw a diameter of the parabola, GH, through the 

M 








178 


CONIC SECTIONS. 



focus F ; GH is the axis of the parabola, B 
and the point V, where the axis cuts the e 
curve, is called the principal vertex of 
the parabola, or simply the vertex. 

It is evident from Def. 1, that the line 
FH is bisected in the point V. 

5. A tangent is a straight line which E 

meets the curve, but, being produced, does 
not cut it. c 

6. An ordinate to a diameter, is a straight line drawn from 
any point of the curve to meet that diameter, and is parallel 
to the tangent at its vertex. 

Thus, let AC be a tangent to the 
parabola at B, the vertex of the di¬ 
ameter BD. From any point E of the 
curve, draw EGH parallel to AC ; j> 
then is EG an ordinate to the diame¬ 
ter BD. 

It is proved in Prop. IX., that EG 
is equal to GH ; hence the entire line 
EH is called a double ordinate. 

7. An abscissa is the part of a diameter intercepted be¬ 
tween its vertex and an ordinate. 



Thus, BG is the abscissa of the diameter BD, correspond¬ 
ing to the ordinate EG. 

8. A subtangent is that part of a diameter intercepted be¬ 
tween a tangent and ordinate to the point of contact. 

Thus, let EL, a tangent to the curve at E, meet the di¬ 
ameter BD in the point L ; then*LG is the subtangent of BD, 
corresponding to the point E. 

9. The parameter of a diameter is the double ordinate 
which passes through the focus. 

Thus, through the focus F, draw IK parallel to the tan¬ 
gent AC; then is IK the parameter of the diameter BD. 

10. The parameter of the axis is called the principal pa¬ 
rameter, or lotus rectum. 

11. A normal is a line drawn perpendicular to a tangent 
from the point of contact, and terminated by the axis.. 

12. A subnormal is the part of the axis 
intercepted between the normal, and the 
corresponding ordinate. 

Thus, let AB be a tangent to the 
parabola at any point A. From A 
draw AC perpendicular to AB; draw, 
also, the ordinate AD. Then AC is the 
normal, and DC is the subnormal cor¬ 
responding to the point A. 










PARABOLA. 


179 


PROPOSITION I. PROBLEM. 

To describe a parabola . 

Let BC be a ruler laid upon a plane, 
and let DEG be a square. Take a 
thread equal in length to EG, and attach B 
one extremity at G, and the other at 
some point as F. Then slide the side 
of the square DE along the ruler BC, 
and, at the same time, keep the thread 
continually tight by means of the pencil 
A; the pencil will describe one part of 
a parabola, of which F is the focus, and q 
BC the directrix. For, in every posi¬ 
tion of the square, 

AF+AG=AE+AG, 
and hence AF=AE ; 

that is, the point A is always equally distant from the focus 
F and directrix BC. 

If the square be turned over, and moved on the other side 
of the point F, the other part of the same parabola may be 
described. 



PROPOSITION IL THEOREM. 

A tangent to the parabola bisects the angle formed at the 
point of contact . by a perpendicular to the directrix, and a line 
drawn to the focus. 

Let A be any point of the parabola 
AY, from which draw the line AF to 
the focus, and AB perpendicular to the 
directrix, and draw AC bisecting the an¬ 
gle BAF; then will AC be a tangent to 
the curve at the point A. 

For, if possible, let the line AC meet 
the curve in some other point as D. 

Join DF, DB, and BF; also, draw DE 
perpendicular to the directrix. 

Since, in the two triangles ACB, ACF, AF is equal to AB 
(Def. I), AC is common to both triangles, and the angle CAB 
is, by supposition, equal to the angle CAF; therefore CB is 
equal to CF, and the angle ACB to the angle ACF. 










180 


CONIC SECTIONS. 


Again, in the two triangles DCB, DCF, because BC is 
equal to CF, the side DC is common to both triangles, and 
the angle DCB is equal to the angle DCF; therefore DB is 
equal to DF. But DF is equal to DE (Def. 1); hence DB 
is equal to DE, which is impossible (Prop. XVII., B. I.). 
Therefore the line AC does not meet the curve in D; and in 
the same manner it may be proved that it does not meet the 
curve in any 6ther point than A ; consequently it is a tangent 
to the parabola. Therefore, a tangent, &c. 

Cor. 1 . Since the angle FAB continually increases as the 
point A moves toward V, and at V becomes equal to two 
right angles, the tangent at the principal vertex is perpendicu¬ 
lar to the axis . The tangent at the vertex V is called the 
vertical tangent . 

Cor. 2. Since an ordinate to any diameter is parallel to 
the tangent at its vertex, an ordinate to the axis is perpen¬ 
dicular to the axis. 


PROPOSITION III. THEOREM. 

The latus rectum is equal to four times the distance from the 
focus to the vertex. 

Let AY B be a parabola, of which F is the 
focus, and V the principal vertex ; then the r 
latus rectum AFB will be equal to four c 
times FV. 

Let CD be the directrix, and let AC be 
drawn perpendicular to it; then, according D 
to Def. 1, AF is equal to AC or DF, because 
ACDF is a parallelogram. But DY is equal 
to YF; that is, DF is equal to twice YF. 

Hence AF is equal to twice YF. In the 
same manner it may be proved that BF is equal to twice 
YF> consequently AB is equal to four times YF, There¬ 
fore, the latus rectum, &c. 



PROPOSITION IV. THEOREM. 

If a tangent to the parabola cut the axis produced , the 
points of contact and of intersection are equally distant from 
the focus. 

Let AB be a tangent to the parabola GAH at.the point A, 
and let it cut the axis produced in B; also, let AF be drawn 
to the focus; then will the line AF be equal to BF. 






PARABOLA. 


181 


Draw AC perpendicular to the di¬ 
rectrix ; then, since AC is parallel to 
BF, the angle BAC is equal to ABF. 

But the angle BAC is equal to BAF 
(Prop. II.) ; hence the angle ABF is 
equal to BAF, and, consequently, AF 
is equal to BF. Therefore, if a tan¬ 
gent, &c. 

Cor. 1 . Let the normal AD be 
drawn. Then, because BAD is a 
right angle, it is equal to the sum of the two angles ABD, 
ADB, or to the sum of the two angles BAF, ADB. Take 
away the common angle BAF, and we have the angle DAF 
equal to ADF. Hence the line AF is equal to FD. There¬ 
fore, if a circle be described with the center F, and radius FA, 
it will pass through the three points B, A, D. 

Cor. 2. The normal bisects the angle made by the diameter 
at the point of contact , with the line drawn from that point to 
the focus. 

For, because BD is parallel to CE, the alternate angles 
ADF, DAE are equal. But the angle ADF has been proved 
equal to DAF; hence the angles DAF, DAE are equal to 
each other. 

Scholium. It is a law in Optics, that the angle made by a 
ray of reflected light with a perpendicular to the reflecting 
surface, is equal to the angle which the incident ray makes 
with the same perpendicular. Hence, if GAH represent a 
concave parabolic mirror, a ray of light falling upon it in the 
direction EA would be reflected to F. The same would be 
true of all rays parallel to the axis. Hence the point F, in 
which all the rays would intersect each other, is called the 
focus. 



PROPOSITION V. THEOREM. 

The subtangent to the axis is bisected by the vertex. 

Let AB be a tangent to the parab¬ 
ola ADV at the point A, and AC 
an ordinate to the axis; then will 
BC be the subtangent, and it will be 
bisected at the vertex V. 

For BF is equal to AF (Prop. 

IV.) ; and AF is equal to CE, which 
is the distance of the point A from 
the directrix. But CE is equal to 
the sum of CV and VE, or CV and VF. Hence BF, or 










182 


CONJC SECTIONS. 


BV+VF, is equal to CV+VF; that is, BY is equal to CY. 
Therefore, the subtangent, &c. 

Cor. 1. Hence the tangent at D, the extremity of the latus 
rectum, meets the axis in E, the same point with the direc¬ 
trix. For, by Def. 8, EF is the subtangent corresponding to 
the tangent DE. 

Cor. 2. Hence, if it is required to draw a tangent to the 
curve at a given point A, draw the ordinate AC to the axis. 
Make BY equal to YC; join the points B, A, and the line 
BA will be the tangent required. 


PROPOSITION VI. THEOREM. 

The subnormal is equal to half the latus rectum. 

Let AB be a tangent to the parab¬ 
ola AY at the point A, let AC be 
the ord-inate, and AD the normal from 
the point of contact; then CD is the 
subnormal, and is equal to half the 
latus rectum. 

For the distance of the point A 
from the focus, is equal to its distance 
from the directrix, which is equal to 
YF+YC, or 2YF+FC ; that is, 

FA=2YF+FC, 
or 2VF=FA—FC. 

Also, CD is equal to FD—FC, which is equal to FA—FC 
(Prop. IV., Cor. 1). Hence CD is equal to 2VF, which is 
equal to half the latus rectum (Prop. III.). Therefore, the 
subnormal, &c. 



PROPOSITION VII. THEOREM. 

If a perpendicular be drawn from the focus to any tangent , 
the point of intersection will be in the vertical tangent . 

Let AB be any tangent to the pa¬ 
rabola AV, and FC a perpendicular let 
fall from the* focus upon AB ; join YC ; 
then will the line VC be a tangent to 
the curve at the vertex Y. 

Draw the ordinate AD to the axis. 

Since FA is equal to FB (Prop. IV.), 
and FC is drawn perpendicular to 
AB, it divides the triangle AFB into 







PARABOLA. 183 

two equal parts, and, therefore, AC is equal to BC. But 
BY is equal to YD (Prop. Y.) ; hence 

BC : CA : : BY : YD, 

and, therefore, CV is parallel to AD (Prop. XYI., B. IY.). But 
AD is perpendicular to the axis BD; hence CY is also per¬ 
pendicular to the axis, and is a tangent to the curve at the 
point Y (Prop. II., Cor. 1). Therefore, if a perpendicular, 
&c. 

Cor . 1 . Because the triangles FYC, FCA are similar, we 
have FY : FC : : FC : FA; 

that is, the perpendicular from the focus upon any tangent , is a 
mean proportional between the distances of the focus from the 
vertex , and from the point of contact. 

Cor. 2. It is obvious that FV : FA : : FC 2 : FA 2 . 

Cor. 3. From Cor. 1, we have 

FC 2 =FYxFA. 

But FY remains constant for the same parabola; therefore 
the distance from the focus to the point of contact , varies as the 
square of the perpendicular-upon the tangent. 


PROPOSITION VIII. THEOREM. 

The square of an ordinate to the axis , is equal to the product 
of the latus rectum by the corresponding abscissa. 

Let AYC be a parabola, and A any point 
of the curve. From A draw the ordinate 
AB; then is the square of AB equal to the 
product of YB by the latus rectum. 

For AB 2 is equal to AF 2 —FB 2 . 

But AF is equal to YB+YF, and FB is 
equal to YB—VF. 

Hence AB 2 = (YB+YF) 3 -(YB—YF) 2 , 
which, according to Prop. IX., Cor., B. IY., 
is equal to 

4YBXVF, 

or YB X the latus rectum (Prop. III.). 

Therefore, the square, &c. 

Cor. 1 . Since the latus rectum is constant for the same 
parabola, the squares of ordinates to the axis, are to each other 
as their corresponding abscissas. 

Cor. 2. The preceding demonstration is equally applicable 
to ordinates on either side of the axis; hence AB is equal to 
BC, and AC is called a double ordinate. The curve is sym¬ 
metrical with respect to the axis, and the whole parabola is 
bisected by the axis. 





184 


CONIC SECTIONS. 


PROPOSITION IX. THEOREM. 

The square of an ordinate to any diameter , is equal to four 
times the product of the corresponding abscissa, by the distance 
from the vertex of that diameter to the focus. 

Let AD be a tangent to the 
parabola VAM at the point 
A; through A draw the di¬ 
ameter HAC, and through h 
any point of the curve, as B, 
draw BC parallel to AD; 
draw also AF to the focus; & 
then will the square of BC be 
equal to 4AF X AC. IC 

Draw CE parallel, and EBG V IT 
perpendicular to the directrix HK; and join BH, BF, HF. 
Also, produce CB to meet HF in L. 

Because the right-angled ’triangles FHK, HCL are simi¬ 
lar, and AD is parallel to CL, we have 

HF:FK::HC: HL 
: : AC : DL. 

Hence (Prop. I., B. II.), 

HFxDL= FK X AC, 

or 2HFxDL=2FKx AC, or 4VFXAC. 

But 2HFxDL=HL 2 —LF 2 (Prop. X., B. IV.) 

= HB 2 —BF 2 
= HG 2 or CE 2 . 

Hence CE 2 is equal to 4VF X AC. 

Also, because the triangles BCE, AFD are similar, we have 
CE : CB : : DF : AF. 

Therefore CE 2 : CB 2 : : DF 2 : AF 2 (Prop. X., B. II.) 

: : VF : AF (Prop. VII., Cor. 2) 

: : 4VFXAC : 4AFxAC. 

But the two antecedents of this proportion have been proved 
to be equal; hence the consequents are equal, or 
BC 2 = 4AFXAC. 

Therefore, the square of an ordinate, &c. 

Cor. In like manner it may be proved that the square of 
CM is equal to 4AFx AC. Hence BC is equal to CM ; and 
since the same may be proved for any ordinate, it follows 
that every diameter bisects its double ordinates. 








PARABOLA. 


185 


PROPOSITION X. THEOREM. 

The parameter of any diameter, is equal to four times the 
distance from its vertex to the focus. 

Let BAD be a parabola, of which 
F is the focus, AC is any diameter, 
and BD its parameter; then is BD . 
equal to four times AF. 

Draw the tangent AE; then, since 
AEFC is a parallelogram, AC is equal E 
to EF, which is equal to AF (Prop. 

IV.). 

Now, by Prop. IX., BC 2 is equal to 
4AFxAC ; that is, to 4AF 2 . Hence 
BC is equal to twice AF, and BD is equal to four times AF. 
Therefore, the parameter of any diameter, &c. 

Cor . Hence the square of an ordinate to a diameter , is 
equal to the product of its parameter by the corresponding 
abscissa. 



PROPOSITION XI. THEOREM. 


If a cone be cut by a plane parallel to its side, the section is 
a parabola. 

Let ABGCD be a cone cut by a plane a 

VDG parallel to the slant side AB; then 
will the section DVG be a parabola. 

Let ABC be a plane section through 
the axis of the cone, and perpendicular to 
the plane VDG; then VE, which is their 
common section, will be parallel to AB. 

Let bgcd be a plane parallel to the base 
of the cone ; the intersection of this plane 
with the cone will be a circle. Since the 
plane ABC divides the cone into two 
equal parts, BC is a diameter of the circle 
BGCD, and be is a diameter of the circle bgcd. Let DEG, 
'deg be the common sections of the plane VDG with the 
planes BGCD, bgcd respectively. Then DG is perpendicular 
to the plane ABC, and, consequently, to the lines VE, BC. 
For the same reason, dg is perpendicular to the two lines 
VE, be. 







186 


CONIC SECTIONS. 


Now, since be is parallel to BE, and 
bB to eE, the figure bBEe is a parallelo¬ 
gram, and be is equal to BE. But be¬ 
cause the triangles Yec, VEC are similar, 
we have 

ec : EC : : Ye : YE ; 

and multiplying the first and second terms 
of this proportion by the equals be and 
BE, we have 

beY.ec : BExEC : :Ve : YE. 

But since be is a diameter of the circle 
bgcd , and de is perpendicular to be (Prop. 

XXII., Cor., B. IV.), 

beYec — de*. ' 

For the same reason, BExEC=DE 2 . 

Substituting these values of beYec and BExEC in the pre¬ 
ceding proportion, we have 

de 2 : DE 3 : : Ye : YE; 

that is, the squares of the ordinates are to each other as the 
corresponding abscissas ; and hence the curve is a parabola, 
whose axis is YE (Prop. YIII., Cor. 1.). Hence the parab¬ 
ola is called a conic section , as mentioned on page 177. 



PROPOSITION XII. THEOREM. 


Every segment of a parabola is two thirds of its circum¬ 
scribing parallelogram. 


K A 


Let AYD be a segment of 
a parabola cut off by the 
straight line AD perpendicu¬ 
lar to the axis; the area of 
AYD is two thirds of the cir¬ 
cumscribing rectangle ABCD. .-r-j. g- 

Draw the line AE touching 
the parabola at A, and meet¬ 
ing the axis produced in E; 
and take a point H in the 

curve, so near to A that the c D 

tangent and curve may be regarded as coinciding. Through 
H draw KL perpendicular, and MN parallel to the axis. 
Then the 

rectangle AL : rectangle AM : : AG X GL : AB x AN 

: : AGxGE : ABxAG 
: : GE : AB, 


m 




H 

E V 

\ 14 











PARABOLA. 


187 


because GL or NH : AN : : GE : AG. But GE is equal to 
twice GY or AB (Prop. Y.) ; hence 

AL:AM : :2: 1 ; 

that is, AL is double of AM. 

Hence the portion of the parabola included between two or¬ 
dinates indefinitely near, is double the corresponding portion 
of the external space ABV. Therefore, since the same is 
true for every point of the curve, the whole space AYG is 
double the space ABY. Whence AVG is two thirds of 
ABYG; and the segment AVD is two thirds of the rectan¬ 
gle ABCD. Therefore, every segment, &c. 


188 


CONIC SECTIONS. 


ELLIPSE. 

Definitions. 

1. An ellipse is a plane curve, in which the sum of the dis¬ 
tances of each point from two fixed points, is equal to a given 
line. 

2. The two fixed points are called the foci. 

Thus, if F, F' are two fixed points, 

and if the point D moves about F in 
such a manner that the sum of its dis¬ 
tances from F and F' is always the 
same, the point D will describe an 
ellipse, of which F and F' are the foci. 

3. The center is the middle point of 
the straight line joining the foci. 

4. The eccentricity is the distance from the center to either 
focus. 

Thus,’ let ABA'B' be an ellipse, 

F and F' the foci. Draw the line 
FF' and bisect it in C. The point 
C is the center of the ellipse ; and 
CF or CF' is the eccentricity. 

5. A diameter is a straight line 
drawn through the center, and 
terminated both ways by the 
curve. 

6. The extremities of a diameter are called its vertices. 

Thus, through C draw any straight line DD' terminated 

by the curve; DD' is a diameter of the ellipse; D and D' 
are its vertices. 

7. The major axis is the diameter which passes through 
the foci; and its extremities are called the principal vertices. 

8. The minor axis is the diameter which is perpendicular 
to the major axis. 

Thus, produce the line FF' to meet the curve in A and 
A'; and through C draw BB' perpendicular to AA'; then is 
AA' the major axis, and BB' the minor axis. 

9. A tangent is a straight line which meets the curve, but, 
being produced, does not cut it. 


B 











ELLIPSE. 


189 


10. An ordinate to a diameter, is a straight line drawn 
from any point of the curve to the diameter, parallel to the 
tangent at one of its vertices. 

Thus, let DD' be any diameter, 
and TT' a tangent to the ellipse 
at D. From any point G of the 
curve draw GKG' parallel to TT' 
and cutting DD' in K; then is 
GK an ordinate to the diameter 
DD'. 

It is proved in Prop. XIX., Cor. 

1, that GK is equal to G'K ; hence the entire line GG' is call¬ 
ed a double ordinate. 

11. The parts into which a diameter is divided by an or¬ 
dinate, are called abscissas. 

Thus, DK and D'K are the abscissas of the diameter DD #jr 
corresponding to the ordinate GK. 

12. The parameter of a diameter, is the double ordinate 
which passes through one of the foci. 

Thus, through the focus F draw r HH' parallel to the tan¬ 
gent TT'; then is HH' the parameter of the diameter DD 7 . 

13. Two diameters are conjugate to one another, when 
each is parallel to the ordinates of the other. 

Thus, draw the diameter EE' parallel to GK, an ordinate 
to the diameter DD', in which case it will, of course, be par¬ 
allel to the tangent TT'; then is the diameter EE' conju¬ 
gate to DD'. * 

14. The parameter of the major axis, is called the princi¬ 
pal parameter, or latus rectum. 

Thus, through the focus F' 
draw LL' a double ordinate to 
the major axis, it will be the latus 
rectum of the ellipse. 

15. A subtangent is that part 
of the axis produced which is in¬ 
tercepted by a tangent, and the 
ordinate drawh from the point of 
contact. 

Thus, if TT' be a tangent to the curve at D, and DG an 
ordinate to the major axis, then GT is the corresponding 
subtangent. 








190 


CONIC SECTIONS. 


PROPOSITION I. PROBLEM. 

* 

To describe an ellipse . 

Let F and F' be any two fixed 
points. Take a thread longer than 
the distance FF', and fasten one of 
its extremities at F, the other at F'. 

Then let a pencil be made to glide 
along the thread so as to keep it al¬ 
ways stretched; the curve described 
by the point of the pencil will be ah 
ellipse. For, in every position of the 
’pencil, the sum of the distances DF, DF' will be the same, 
viz., equal to the entire length of the string. 



PROPOSITION II. THEOREM. 


The sum of the two lines drawn from any point of an ellipse 
to the foci , is equal to the major axis . 



Let ADA' be an ellipse, of 
which F, F' are the foci, AA' is 
the major axis, and D any point of 
the curve; then will DF+DF' be £ 
equal to AA'. 

For, by Def. 1, the sum of the 
distances of any point of the curve 
from the foci, is equal to a given line. Now, when the point 
D arrives at A, FA+F'A or 2AF+FF' is equal to the given 
line. And when D is at A', FA'+F'A' or 2A'F'+FF' is 
equal to the same line. Hence 


2AF+FF'=2A'F'+FF'; 
consequently, AF is equal to A'F'. 

Hence DF+DF', which is equal to AF+AF', must be equal 
to AA'. Therefore, the sum of the two lines, &c. 

Cor. The major axis is bisected in the center. For, by Def. 
3, CF is equal to CF'; and we have just proved that AF is 
equal to A'F'; therefore AC is equal to A'C. 





ELLIPSE. 


191 


PROPOSITION III. THEOREM. 

Every diameter is bisected in the center . 

Let D be any point of an ellipse; 
join DF, DF', and FF'. Complete the 
parallelogram DFD'F', and join DD'. 

Now, because the opposite sides of 
a parallelogram are equal, the sum of 
DF and DF' is equal to the sum of 
D'F and D'F'; hence D' is a point in 
the ellipse. But the diagonals of a parallelogram bisect each 
other; therefore FF' is bisected in C ; that is, C is the center 
of the ellipse, and DD' is a diameter bisected in C. There¬ 
fore, every diameter, &c. 



PROPOSITION IV. THEOREM. 

The distance from either focus to the extremity of the minor 
axis , is equal to half the major axis . 

Let F and F' be the foci of an B 

ellipse, AA' the major axis, and 
BB' the minor axis; draw the 
straight lines BF, BF'; then BF, ^ 

BF' are each equal to AC. 

In the two right-angled trian¬ 
gles BCF, BCF', CF is equal to 
CF', and BC is common to both B' 

triangles ; hence BF is equal to BF'. But BF+BF' is equal 
to 2AC (Prop. II.); consequently, BF and BF' are each 
equal to AC. Therefore, the distance, &c. 

Cor. 1. Half the minor axis is a mean proportional between 
the distances from either focus to the principal vertices. 

For BC 2 is equal to BF 2 —FC a (Pro.p. XI., B. IV.), which 
is equal to AC 2 —FC 2 (Prop. IV.). Hence (Prop. X., B. IV.), 
BC 2 =(AC+FC) x (AC-FC) 

= AF' x AF ; and, therefore, 

AF : BC : : BC : FA'. 

Cor. 2. The square of the eccentricity is equal to the differ¬ 
ence of the squares of the semi-axes. 

For FC 2 is equal to BF 2 —BC 2 , which is equal to AC 3 — 
BC 2 . 






192 


CONIC SECTIONS. 


PROPOSITION V. THEOREM. 

A tangent to the ellipse . makes equal angles with straight 
lines drawn from the point of contact to the foci. 

Let F, F' be the foci of an ellipse, 
and D any point of the curve; if 
through the point D the line TT' 
be drawn, making the angle TDF 
equal to T'DF', then will TT' be 
a tangent to the ellipse at D. 

For if TT' be not a tangent, it 
must meet the curve in some other 
point than D. Suppose it to meet the curve in the point E. 
Produce F'D to G, making DG equal to DF; and join EF, 
EF', EG, and FG. 

Now, in the two triangles DFH, DGH, because DF is equal 
to DG, DH is common to both triangles, and the angle FDH 
is, by supposition, equal to F'DT', which is equal to the ver¬ 
tical angle GDH; therefore HF is equal to HG, and the an¬ 
gle DHF is equal to the angle DHG. Hence the line TT' 
is perpendicular to FG at its middle point; and, therefore, 
EF is equal to EG. 

Also, F'G is equal to F'D+DF, or F'E+EF, from the na¬ 
ture of the ellipse. But F'E+EG is greater than F'G (Prop. 
VIII., B. I.); it is, therefore, greater than F'E+EF. Con¬ 
sequently EG is greater than EF; which is impossible, for 
we have just proved EG equal to EF. Therefore E is not 
a point of the curve, and TT' can not meet the curve in any 
other point than D; hence it is a tangent to the curve at the 
point D. Therefore, a tangent to the ellipse, &c. 

Cor. 1 . The tangents at the vertices of the axes, are per¬ 
pendicular to the axes; and hence an ordinate to either axis 
is perpendicular to that axis. 

Cor. 2. If TT' represent a plane mirror, a ray of light 
proceeding from F in the direction FD, would be reflected in 
the direction DF', making the angle of reflection equal to the 
angle of incidence. And, since the ellipse may be regarded 
as coinciding with a tangent at the point of contact, if rays 
of light proceed from one focus of a concave ellipsoidal mir¬ 
ror, they will all be reflected to the other focus. For this 
reason, the points F, F' are called the foci, or burning points. 





ELLIPSE. 


193 


PROPOSITION VI. THEOREM. 


Tangents to the ellipse at the vertices of a diameter , are par - 
allel to each other. 

Let DD' be any diameter 
of any ellipse, and TT', 

VV' tangents to the curve 
at the points D, D'; then 
will they be parallel to each , 
other. 

Join DF, DF', D'F, D'F'; 
then, by the preceding Prop¬ 
osition, the angle FDT is 
equal to F'DT', and the an¬ 
gle FD'V is equal to F'D'V'. But, by Prop. III., DFD'F' is 
a parallelogram ; and since the opposite angles of a parallelo¬ 
gram are equal, the angle FDF' is equal to FD'F'; therefore 
the angle FDT is equal to F'D'V' (Prop. II., B. I.). Also, 
since FD is parallel to F'D', the angle FDD' is equal to 
F'D'D; hence the whole angle D'DT is equal to DD'V'; 
and, consequently, TT' is parallel to VV'. Therefore, tan¬ 
gents, &c. 

Cor. If tangents are drawn through the vertices of any 
two diameters, they will form a parallelogram circumscribing 
the ellipse. 



PROPOSITION VII. THEOREM. 


If from the vertex of any diameter , straight lines are drawn 
through the foci , meeting the conjugate diameter , the part in¬ 
tercepted by the conjugate is equal to half the major axis. 


Let EE' be a diameter conju¬ 
gate to DD', and let the lines DF, 

DF' be drawn, and produced, if 
necessary, so as to meet EE' in 
H and K; then will DH or DK 
be equal to AC. 

Draw FG parallel to EE' or 
TT'. Then the angle DGF is 
equal to the alternate angle 
F'DT', and the angle DFG is equal to FDT. But the angles 
FDT, F'DT' are equal to each other (Prop. V.); hence the 

N 







194 


CONIC SECTIONS. 


angles DGF, DFG are equal to each other, and DG is equal 
to DF. Also, because CH is parallel to FG, and CF is equal 
to CF'; therefore HG must be equal to HF'. 

Hence FD-fF'D is equal to 2DG+2GH or 2DH. But 
FD+F'D is equal to 2AQ. Therefore 2AC is equal to 2DH, 
or AC is equal to DH. 

Also, the angle DHK is equal to DKH ; and hence DK is 
equal to DH or AC. Therefore, if from the vertex, &c. 


PROPOSITION VIII. THEOREM. 


Perpendiculars drawn from the foci upon a tangent to the 
ellipse , meet the tangent in the circumference of a circle , whose 
diameter is the major axis. 


AG- 


Let >TT' be a tangent to the 
ellipse at D, and from F' draw 
F'E perpendicular to T'T ; the 
point E will be in the circum¬ 
ference of a circle described 
upon AA' as a diameter. 

Join CE, FD, F'D, and pro¬ 
duce F'E to meet FD produced 
in G. 

Then, in the two triangles 
DEF', DEG, because DE is com¬ 
mon to both triangles, the angles # 

at E are equal, being right angles; also, the angle EDF' is 
equal to FDT (Prop. V.), which is equal to the vertical an¬ 
gle EDG; therefore DF' is equal to DG, and EF' is equal 
to EG. 4 

Also, because F'E is equal to EG, and F'C is equal to CF, 
CE must be parallel to FG, and, consequently, equal to half 
of FG. 



But, since DG has been proved equal to DF', FG is equal 
to FD+DF', which is equal to AA'. Hence CE is equal to 
half of AA' or AC ; and a circle described with C as a cen¬ 
ter, and radius CA, will pass through the point E. The same 
may be proved of a perpendicular let fall upon TT' from the 
focus F. Therefore, perpendiculars, &c. 



ELLIPSE. 


195 


PROPOSITION IX. THEOREM. 

The product of the perpendiculars from the foci upon a tan¬ 
gent , is equal to the square of half the minor axis . 

Let TT' be a tangent to the 
ellipse at any point E, and let 
the perpendiculars FD, F'G be 
drawn from the foci; then will 
the product of FD by F'G, be 
equal to the square of BC. ^ 

On AA', as a diameter, de¬ 
scribe a circle ; it will pass 
through the points D and G 
(Prop. VIII.). Join CD, and 
produce it to meet GF' in D'. 

Then, because FD and F'G are perpendicular to the same 
straight line TT', they are parallel to each other, and the al¬ 
ternate angles CFD, CF'D' are equal. Also, the vertical 
angles DCF, D'CF' are equal, and CF is equal to CF'. 
Therefore (Prop. VII., B. I.) DF is equal to D'F', and CD is 
equal to CD'; that is, the point D' is in the circumference of 
the circle ADGA'. 

Hence DFxGF' is equal to D'F'xGF', which is equal to 
A'F'xF'A (Prop. XXVII., B. IV.), which is equal to BC a 
(Prop. IV., Cor. 1). Therefore, the product, &c. 

Cor. The triangles FDE, F'GE are similar ; hence 
FD : F'G : : FE : F'E; 

that is, perpendiculars let fall from the foci upon a tangent , 
are to each other as the distances of the point of contact from 
the foci. 



PROPOSITION X. THEOREM. 

If a tangent and ordinate he drawn from the same point of 
an ellipse , meeting either axis produced , half of that axis will 
be a mean proportional between the distances of the two inter¬ 
sections from the center. 

Let TT' be a tangent to the ellipse, and DG an ordinate 
to the major axis from the point of contact; then we shall 
have GT ; CA . • CA : CG. 

Join DF, DF'; then, since the exterior angle of the trian¬ 
gle FDF' is bisected by DT (Prop. V.), we have 








196 


CONIC SECTIONS. 


T' 


T 


F'T : FT : : F'D : FD (Prop. XVII., Sch., B. IY.). 
Hence, by Prop. YII, Cor., B. II., 

F'T+FT : F'T —FT : : F'D+FD : F'D—FD, 
or 2CT : F'F : : 2CA : F'D—FD ; 

that is, 2CT : 2CA : ; F'F : F'D—FD. (1) 

Again, because DG is drawn from the vertex of the trian¬ 
gle FDF' perpendicular to the base FF', we have (Prop. 
XXXI., Cor., B. IY.), 

F'F : F'D—FD : : F'D+FD : F'G—FG, 
or F'F : F'D-FD : : 2CA : 2CG. (2) 

Comparing proportions (1) and (2), we have 
2CT : 2CA : : 2CA : 2CG, 
or CT: CA : : CA : CG. 

It may also be proved that 

CT' : CB : : CB : CG'. 

Therefore, if a tangent, &c. 


PROPOSITION XI. THEOREM. 

The suhtangent of an ellipse , is equal to the corresponding 
subtangent of the circle described upon its major axis. 

Let AEA' be a circle de¬ 
scribed on A A', the major 
axis of an ellipse; and from 
any point E in the circle, 
draw the ordinate EG meet- & 
ing the ellipse in D. Draw 
DT touching the ellipse at 
D; join ET; then will ET 
be a tangent to the circle at E. 

Join CE. Then, by the last Proposition, 

CT : CA : : CA : CG; 
or, because CA is equal to CE, 

CT : CE : : CE : CG. 

Hence the triangles CET, CGE, having the angle at C com¬ 
mon, and the sides about this angle proportional, are similar. 
Therefore the angle CET, being equal to the angle CGE, is 











ELLIPSE. 


197 


a right angle; that is, the line ET is perpendicular to the 
radius CE, and is, consequently, a tangent to the circle (Prop. 
IX., B. III.). Hence GT is the subtangent corresponding to 
each of the tangents DT and ET. Therefore, the subtan¬ 
gent, &c. 

Cor. A similar property may be proved of a tangent to the 
ellipse meeting the minor axis. 


PROPOSITION XII. THEOREM. 

The square of either axis, is to the square of the other, as the 
rectangle of the abscissas of the former, is to the square of their 
ordinate. 


Let DE be an ordinate to the 
major axis from the point D; 
then we shall have 
CA 2 : CB 2 : : AE xEA': DE 2 . 

Draw TT' a tangent to the 
ellipse at D, then, by Prop. X., 

CT : CA : : CA : CE. 

Hence (Prop. XII., B. II.), 

CA 2 : CE 2 : : CT : CE; 
and, by division (Prop. VII., B. 

II.), CA 2 : CA 3 —CE 2 

Again, by Prop. X., 

CT' : CB : : CB : CE' or DE. 
Hence (Prop. XII., B. II.), 

CB 2 : DE 2 : : CT' : DE. 



CT: ET. 


But, by similar triangles, 

CT' : DE : : CT : ET; 

therefore CB 2 : DE 2 : : CT : ET. (2) 

Comparing proportions (1) and (2), we have 
CA 2 : CA 2 —CE 2 : : CB 2 : DE 2 . 


But CA 2 —CE 2 is equal to AE X EA' (Prop. X., B. IY.); 
hence CA 2 : CB 3 : : AE X EA' : DE 2 . 

In the same manner it may be proved that 

CB 2 : CA 2 : : BE'xE'B' : DE' 2 . 


Therefore, the square, &c. 

Cor. 1 . CA 2 : CB 2 : : CA 2 -CE 2 : DE 2 . 

Cor. 2. The squares of the ordinates to either axis, are to 
each other as the rectangles of their abscissas. 

Cor. 3. If a circle be described on either axis, then any or¬ 
dinate in the circle, is to the corresponding ordinate in the 
ellipse, as the axis of that ordinate, is to the other axis . 








198 


CONIC SECTIONS. 


For, by the Proposition, 

CA 2 : CB 2 : : AE xEA' : DE 2 . 

But AE X EA' is equal to GE 2 
(Prop. XXII., Cor., B. IV.). 

Therefore CA 2 : CB 2 : : GE 2 : DE 2 , X 
or CA: CB : : GE : DE. 

In the same manner it may be 
proved that 

CB : CA : : G'E' : DE'. 



PROPOSITION XIII. THEOREM. 

The latus rectum is a third proportional to the major and 
minor axes, B 

Let LL' be a double ordinate to 
the major axis passing through the 
focus F; then we shall have 
AA' : BB' : : BB' : LL'. 

Because LF is an ordinate to the 
major axis, B' 

AC 2 : BC 2 : : AFxFA' : LF 2 (Prop. XII.). 

: : BC 2 : LF 2 (Prop. IV., Cor. 1). 

Hence AC : BC : : BC : LF, 

or AA': BB' : : BB' : LL'. 

Therefore, the latus rectum, &c. 



PROPOSITION XIV. THEOREM. 

If from the vertices of two conjugate diameters, ordinates are 
drawn to either axis , the sum of their squares will he equal to 
the square of half the other axis. 

Let DD', EE' be any 
two conjugate diameters, 

DG and EH ordinates to 
the major axis drawn 
from their vertices; in T 
which case, CG and CH 
will be equal to the ordi¬ 
nates to the minor axis 
drawn from the same points ; then we shall have 
CA 2 = CG 2 +CH 2 , and CB 2 =DG 2 +EH 2 . 

Let DT be a tangent to the ellipse at D, and ET' a tan¬ 
gent at E. Then, by Prop. X., 

















ELLIPSE. 


199 


CGxCT is equal to CA 2 , or CHxCT'; 
whence CG : CH : : CT' : CT ; or, by similar triangles, 

: : CE : DT ; that is, 

: : CH : GT. 

Hence CH 2 = GT xCG, 

= (CT —CG) xCG 
= CG xCT-CG 2 
= CA 2 —CG 2 (Prop. X,) ; 
that is, CA 2 = CG 2 +CH 2 . 

In the same manner it may be proved that 

CB 2 = DG 2 +EH 2 . 

Therefore, if from the vertices, &c. 

Cor . 1 . CH 2 is equal to CA 2 — CG 2 ; that is, CGxGT; 
hence (Prop. XII., Cor. 1), 

CA 2 : CB 2 :: CGxGT : DG 2 . 

Cor. 2. CG 2 is equal to CA 2 -CH 2 or AH x HA' ; hence 
CA 2 : CB 2 : : CG 2 : EH 2 . 


PROPOSITION XV. THEOREM. 


The sum of the squares of any two conjugate diameters , is 
equal to the sum of the squares of the axes. 

Let DIP, EE' be any two con¬ 
jugate diameters; then we shall 
have 

DD' 2 +EE' 2 =AA' 2 -fBB' 2 . 

Draw DG, EH ordinates to the 
major axis. Then, by the prece¬ 
ding Proposition, 

CG 2 +CH 2 =CA 2 , 
and DG 2 +EH 2 = CB 2 . 

Hence CG 2 +DG 2 +CH 2 -f EH 2 = CA 2 +CB 2 , 

or CD 2J rCE 2 = CA 2 -t-CB 2 ; 

that is, DD' 2 +EE' 2 =AA' 2 -fBB' 2 . 

Therefore, the sum of the squares, &c. 



PROPOSITION XVI. THEOREM. 

The parallelogram formed by drawing tangents through the 
vertices of two conjugate diameters , is equal to the rectangle of 
the axes. 

Let DED'E' be a parallelogram, formed by drawing tan¬ 
gents to the ellipse through the vertices of two conjugate 
diameters DD', EE'; its area is equal to AA'xBB'. 









CONIC SECTIONS. 


200 



Let the tangent at D, meet the major axis produced in T; 
join E'T, and draw the ordinates DG, E'H. 

Then, by Prop. XIV., Cor. 2, we have 
CA 2 : CB 2 : : CG 2 : E'H 2 , 
or CA : CB : : CG : E'H. 

But CT : CA : : CA : CG (Prop. X.) ; 

hence CT : CB : : CA : E'H, 

or CA x CB is equal to CT x E'H, 

which is equal to twice the triangle CE'T, or the parallelo¬ 
gram DE'; since the triangle and parallelogram have the 
same base CE', and are between the same parallels. 

Hence 4CAxCB or AA'xBB', is equal to 4DE', or the 
parallelogram DED'E'. Therefore, the parallelogram, &c. 


PROPOSITION XVII. THEOREM. 


If from the vertex of any diameter , straight lines are drawn 
to the foci, their product is equal to the square of half the con¬ 
jugate diameter. 


Let DD', EE' be two conjugate ^ j£ 
diameters, and from D let lines 
be drawn to the foci; then will 
FDxF'D be equal to EC 2 . 

Draw a tangent to the ellipse 
at D, and upon it let fall the per- A( 
pendiculars FG, F'H ; draw, also, 

DK perpendicular to EE'. 

Then, because the triangles 
DFG, DLK, DF'H are similar, we have 
FD : FG : : DL : DK. 

Also, F'D : F'H : : DL : DK. 

Whence (Prop. XI., B. II.), 

FD x F'D : FG X F'H : : DL 2 : DK 2 . 
But, by Prop. XVI., AC x BC = EC x DK; 
whence AC or DL : DK : : EC : BC, 

and DL 2 : DK 2 : : EC 2 : BC 2 . 


f 1 

/f\K; 

■If / 

iy 



( 1 ) 


(2) 










ELLIPSE. 


201 


Comparing proportions (1) and (2), we have 

FD x F'D : FG x Fid : : EC 2 : BC 2 . 

But FGxFTI is equal to BC 2 (Prop. IX.) ; hence FDxF'D 
is equal to EC 2 . Therefore, if from the vertex, &c. 


PROPOSITION XVIII. THEOREM. 

If a tangent and ordinate be drawn from the same point of 
an ellipse to any diameter , half of that diameter will be a mean 
proportional between the distances of the two intersections from 
the center. 

Let a tangent EG and an ordinate EH be drawn from the 
same point E of an ellipse, meeting the diameter CD pro¬ 
duced ; then we shall have 

CG : CD : : CD : CH. 



Produce EG and EH to meet the major axis in K and L; 
draw DT a tangent to the curve at the point D, and draw 
DM parallel to GK. Also, draw the ordinates EN, DO. 

By Prop. XIV., Cor. 1, CA 2 : CB 2 : : COxOT : DO 2 , 

: : CN x NK : EN 2 . 

Hence 

COxOT : CNxNK : : DO 2 : EN 2 

: : OT 2 : NL 2 , by similar triangles. (1) 
Also, by similar triangles, OT : NL : : DO : EN 

: : OM : NK. (2) 
Multiplying together proportions (1) and (2) (Prop. XI., 
B. II.), and omitting the factor OT 2 in the antecedents, and 
NKxNL in the consequents, we have 

CO : CN : : OM : NL; 

and, by composition, CO : CN : : CM : CL. (3) 

Also, by Prop. X., CKxCN=CA 2 = CT xCO; 
hence CO : CN : : CK : CT. (4) 

Comparing proportions (3) and (4), we have 
CK : CM : : CT : CL. 

But CK : CM : : CG : CD, 

and CT : CL : : CD : CH; 

hence CG : CD : : CD : CH. 

Therefore, if a tangent, &c. 






202 


CONIC SECTIONS. 


PROPOSITION XIX. THEOREM. 

The square of any diameter , is to the square of its conjugate . 
as the rectangle of its abscissas , is to the square of their or¬ 
dinate. 

Let DD', EE' be two conjugate 
diameters, and GH an ordinate to 
DD'; then 

DD' 2 : EE' 2 : : DHxHD' : GH 2 . 

Draw TT' a tangent to the 
curve at the point G, and draw 
GK an ordinate to EE'. Then, 
by Prop. XVIII., 

CT : CD : : CD : CH, 

and CD 2 : CH 2 : : CT : CH (Prop. XII., B. II.); 

whence, by division, 

CD 2 : CD 2 —CH 2 : : CT : PIT. (1) 

Also, by Prop. XVIII., 

CT' : CE : : CE : CK, 

and CE 2 : CK 2 : : CT': CK or GH, 

: : CT : HT. (2) 

Comparing proportions (1) and (2), we have 

CD 2 : CE 2 : : CD 2 -CH 2 : CK 2 or GH 2 , 
or DD /2 : EE' 2 : : DH x HD' : GH 2 . 

Therefore, the square, &c. 

Cor. 1. In the same manner it may be proved that 
DD' 2 : EE' 2 :: DHxHD' : G'H 2 ; hence GH is equal to G'H, 
or every diameter bisects its double ordinates. 

Cor. 2. The squares of the ordinates to any diameter, are 
to each other as the rectangles of their abscissas. 



PROPOSITION XX. THEOREM. 

If a cone be cut by a plane , making ari angle with the base 
less than that made by the side of the cone , the section is an 
ellipse. 

Let ABC be a cone cut by a plane DEGH, making an an¬ 
gle with the base, less than that made by the side of the cone ; 
the section DeEGH/i is an ellipse. 

Let ABC be a section through the axis of the cone, and 
perpendicular to the plane DEGH. Let EMHO, emho be 
circular sections parallel to the base ; then EH, the intersec- 




ELLIPSE. 


203 


tion of the planes DEGH, EMHO, will 
be perpendicular to the plane ABC, 
and, consequently, to each of the lines 
DG, MO. So, also, eh will be perpen¬ 
dicular to DG and mo. 

Now, because the triangles DNO, 

D no are similar, as also the triangles 
GMN, Gmn , we have the proportions, 

NO : no : : DN : D n, 

and MN : mn : : NG : nG. -- 

Hence, by Prop. XI., B. II., 

MN x NO : mn x no : : DN x NG : Dn x nG. 

But since MO is a diameter of the circle EMHO, and EN is 
perpendicular to MO, we have (Prop. XXII., Cor., B. IV.), 
MNxNO=EN 2 . 

For the same reason, mnxno = en 2 . 

Substituting these values of MNxNO and mnxno , in the 
preceding pr6portion, we have 

EN 2 : erf : : DNxNG : DnxnG; 
that is, the squares of the ordinates to the diameter DG, are 
to each other as the products of the corresponding abscissas. 
Therefore the curve is an ellipse (Prop. XII.. Cor. 2) whose 
major axis is DG. Hence the ellipse is called a conic section , 
as mentioned on page 177. 



PROPOSITION XXI. THEOREM. 

The area of an ellipse is a mean proportional between the 
two circles described on its axes. 

Let AA' be the major axis of an 
ellipse ABA'B'. On AA' as a di¬ 
ameter, describe a circle; inscribe 
in the circle any regular polygon 
AEDA', and from the vertices E, 

D, &c., of the polygon, draw per¬ 
pendiculars to AA r . Join the 
points B, G, &c., in w r hich these 
perpendiculars intersect the ellipse, 
and there will be inscribed in the 
ellipse a polygon of an equal num¬ 
ber of sides. 

Now the area of the trapezoid CEDH, is equal to (CE-f- 

fTT 

DH) x-; and the area of the trapezoid CBGH, is equal to 

2 














204 


CONIC SECTIONS. 


(CB -{-GH) x-. These trapezoids 

2 

are to each other, as CE+DH to 
CB+GH, or as AC to BC (Prop. 

XII., Cor. 3). 

In the same-manner it may be A 
proved thaJ; each of the trapezoids 
composing the polygon inscribed in 
the circle, is to the corresponding 
trapezoid of the polygon inscribed 
in the ellipse, as AC to BC. Hence, 
the entire polygon inscribed in the circle, is to the polygon in¬ 
scribed in the ellipse, as AC to BC. 

Since this proportion is true, whatever be the number of 
sides of the polygons, it will be true when the number is in¬ 
definitely increased ; in which case one of the polygons coin¬ 
cides with the circle, and the other with the ellipse. Hence 
we have 

Area of circle : area of ellipse : : AC : BC. 

But the area of the circle is represented by ttAC 2 ; hence 
the area of the ellipse is equal to uAC x BC, which is a mean 
proportional between the two circles described on the axes. 







HYPERBOLA. 

Definitions. 

1. An hyperbola is a plane curve, in which the difference 
of the distances of each point from two fixed points, is equal 
to a given line. 

2. The two fixed points are called the foci. 

Thus, if F and F' are two fixed 

points, and if the point D moves 
about F in such a manner that the 
difference of its distances from F and 
F' is always the same, the point D 
will describe an hyperbola, of which 
F and F' are the foci. 

If the point D' moves about F' in 
such a manner that D'F—IFF' is 
always equal to DF'—DF, the point D ; will describe a sec¬ 
ond hyperbola similar to the first. The two curves are call¬ 
ed opposite hyperbolas. 

3. The center is the middle point of the straight line join¬ 
ing the foci. 

4. The eccentricity is the distance from the center to either 
focus. 

Thus, let F and F' be the foci of 
two opposite hyperbolas. Draw the 
line FF', and bisect it in C. The 
point C is the center of the hyperbola, 
and CF or CF' is the eccentricity. 

5. A diameter is a straight line 
drawn through the center, and termi¬ 
nated by two opposite hyperbolas. 

6. The extremities of a diameter are 
called its vertices. 

Thus, through C draw any straight line DD' terminated 
by the opposite curves ; DD' is a diameter of the hyperbola; 
D and D' are its vertices. 

7. The major axis is the diameter which, when produced, 
passes through the foci; and its extremities are called the 
principal vertices. 

8. The minor axis is a line drawn through the center per- 







206 


CONIC SECTIONS. 


pendicular to the major axis, and terminated by the circum¬ 
ference described from one of the principal vertices as a cen¬ 
ter, and a radius equal to the eccentricity. 

Thus, through C draw BB' perpendicular to AA', and with 
A as a center, and with CF as a radius, describe a circum¬ 
ference cutting this perpendicular in B and B'; then AA' is 
the major axis, and BB' the minor axis. 

If on BB' as a major axis, opposite hyperbolas are de¬ 
scribed, having AA' as their minor axis, these hyperbolas are 
said to be conjugate to the former. 

9. A tangent is a straight line which meets the curve, but, 
being produced, does not cut it. 

10'. An ordinate to a diameter, is a straight line drawn 
from any point of the curve to meet the diameter produced, 
parallel to the tangent at one of its vertices. 

Thus, let DD' be any diame¬ 
ter, and TT' a tangent to the 
hyperbola at D. From any 
point G of the curve draw 
GKG' parallel to TT' and cut¬ 
ting DIF produced in K; then 
is GK an ordinate to the di¬ 
ameter DD'. 

It is proved, in Prop. XIX., 

Cor. 1, that GK is equal to 

G'K; hence the entire line GG' is called a double ordinate. 



11. The parts of the diameter produced, intercepted be¬ 
tween its vertices and an ordinate, are called its abscissas. 

Thus, DK and D'K are the abscissas of the diameter DIP 
corresponding to the ordinate GK. 

12. The parameter of a diameter, is the double ordinate 
which passes through one of the foci. 

Thus, through the focus F draw HIP parallel to the tan¬ 
gent TT'; then is HH' the parameter of the diameter DD'. 

13. Two diameters are conjugate to one another, when 
each is parallel to the ordinates of the other. 

Thus, draw the diameter EE' parallel to GK an ordinate 
. to the diameter DD', in which case it will, of course, be par¬ 
allel to the tangent TT'; then is 
the diameter EE' conjugate to DD'. 

14. The parameter of the major 
axis, is called the principal parame¬ 
ter, or latus rectum. 

Thus, through the focus F' draw 
LL' a double ordinate to the major 
axis, it will be the latus rectum of 
the hyperbola. 



* 







HYPERBOLA. 


207 


15. A subtangent is that part of the axis produced which 
is intercepted by a tangent, and the ordinate drawn from the 
point of contact. , 

Thus, if TT' be a tangent to the curve at D, and DG an 
ordinate to the major axis, then GT is the corresponding 
subtangent. 


PROPOSITION I. PROBLEM. 

To describe an hyperbola. 

Let F and F' be any two fixed 
points. Take a ruler longer than 
the distance FF', and fasten one 
of its extremities at the point F'. 

Take a thread shorter than the 
ruler, and fasten one end of it at 
F, and the other to the end H of the 
ruler. Then move the ruler HDF' 
about the point F', while the thread is kept constantly stretched 
by a pencil pressed against the ruler; the curve described by 
the point of the pencil, will be a portion of an hyperbola. 
For, in every position of the ruler, the difference of the lines 
DF, DF' will be the same, viz., the difference between the 
length of the ruler and the length of the string. 

If the ruler be turned, and move on the other side of the 
point F, the other part of the same hyperbola may be de¬ 
scribed. Also, if one end of the ruler be fixed in F, and that 
of the thread in F', the opposite hyperbola may be described. 



PROPOSITION II. THEOREM. 

The difference of the two lines drawn from any point of an 
hyperbola to the foci, is equal to the major axis. 

Let F and F' be the foci of two 
opposite hyperbolas, AA' the major 
axis, and D any point of the curve; 
then will DF' —DF be equal to A A'. 

For, by Def. 1, the difference of the 
distances of any point of the curve 
from the foci, is equal to a given line. 

Now when the point D arrives at A, 

F'A—FA, or AA'+F'A'—FA, is equal to the given line. 
And when D is at A', FA'—F'A', or AA'+AF—A'F', is 
equal to the same line. Hence 






208 


CONIC SECTIONS. 


AA'+AF-A'F'=AA'-f-F'A'-FA, 
or 2AF=2A'F'; 

that is, AF is equal to A'F'. 

Hence DF' —DF, which is equal to AF'—AF, must be 
equal to AA'. Therefore, the difference of the two lines, &c. 

Cor. The major axis is bisected in the center. For, by 
Def. 3, CF is equal to CF'; and we have just proved that 
AF is equal to A'F'; therefore AC is equal to A'C. 


PROPOSITION III. THEOREM. 

Every diameter is bisected in the center. 

Let D be any point of an hyper¬ 
bola ; join DF, DF', and FF'. Com¬ 
plete the parallelogram DFD'F', and 
join DD'. 

Now, because the opposite sides of 
a parallelogram are equal, the differ¬ 
ence between DF and DF' is equal 
to the difference between D'F and 
D'F'; hence D' is a point in the opposite hyperbola. But 
the diagonals of a parallelogram bisect each other; there¬ 
fore FF' is bisected in C; that is, C is the center of the hy¬ 
perbola, and DD' is a diameter bisected in C. Therefore, 
every diameter, &c. - 



PROPOSITION IV. THEOREM. 

Half the minor axis is a mean proportional between the dis¬ 
tances from either focus to the principal vertices. 

Let F and F' be the foci of opposite ^ 
hyperbolas, AA' the major axis, and BB' \ 
the minor axis ; then will BC be a mean \ 
proportional between AF and A'F. T'-Lr 

Join AB. Now BC 2 is equal to AB 2 — / 

AC 2 , which is equal to FC 2 —AC 2 (Def. / 

8). Hence (Prop. X., B. IV.), ^ 

BC 2 =(FC-AC) x (FC+AC) 

= AFx A'F ; 

and hence AF : BC : : BC : A'F. 

Cor. The square of the eccentricity is equal to the sum of the 
squares of the semi-axes. 

For FC 2 is equal to AB 2 (Def. 8), which is equal to AC 2 + 
BC 2 . 






HYPERBOLA. 


209 


PROPOSITION V. THEOREM. 

A tangent to the hyperbola bisects the angle contained by 
lines drawn from the point of contact to the foci. 

Let F, F' be the foci of two 
opposite hyperbolas, and D any 
point of the curve ; if through the 
point D, the line TT' be drawn 
bisecting the angle FDF'; then 
will TT' be a tangent to the hy¬ 
perbola at D. 

For if TT' be not a tangent, let 
it meet the curve in some other 
point, as E. Take DG equal to 
DF; and join EF, EF', EG, and FG. 

Now, in the two triangles DFH, DGH, because DF is 
equal to DG, DH is common to both triangles, and the angle 
FDH is, by supposition, equal to GDH; therefore HF is 
equal to HG, and the angle DHF is equal to the angle DHG. 
Hence the line TT' is perpendicular to FG at its middle 
point; and, therefore, EF is equal to EG. 

Now F'G is equal to F'D —DF, or F'E —EF, from the 
nature of the hyperbola. But F'E —EG is less than F'E 
(Prop. VIII., B. I.); it is, therefore, less than F'E—EF. 
Consequently, EG is greater than EF, which is impossible, 
for we have just proved EG equal to EF. Therefore E is 
not a point of the curve ; and TT' can not meet the curve in 
any other point than D; hence it is a tangent to the curve 
at the point D. Therefore, a tangent to the hyperbola, &c. 

Cor. 1 . The tangents at the vertices of the axes, are per¬ 
pendicular to the axes ; and hence an ordinate to either axis 
is perpendicular to that axis. 

Cor. 2. If TT' represent a plane mirror, a ray of light 
proceeding from F in the direction FD, would be reflected in 
a line which, if produced, would pass through F', making the 
angle of reflection equal to the angle of incidence. And, 
since the hyperbola may be regarded as coinciding with a 
tangent at the point of contact, if rays of light proceed from 
one focus of a concave hyperbolic mirror, they will be re¬ 
flected in lines diverging from the other focus. For this 
reason, the points F, F' are called the foci. 





210 


CONIC SECTIONS. 


PROPOSITION VI. THEOREM. 

Tangents to the hyperbola at the vertices of a diameter, are 
parallel to each other. 

Let DD' be any diameter of an 
hyperbola, and TT', VV' tangents 
to the curve at the points D, I)'; 
then will they be parallel to each 
other. 

Join DF, DF', D'F, D'F'. Then, 
by Prop. III., FDF'D' is a paral¬ 
lelogram; and, since the opposite 
angles of a parallelogram are equal, 
the angle FDF' is equal to FD'F'. 

But the tangents TT', VV' bisect the angles at D and D' 
(Prop. V.); hence the angle F'DT', or its alternate angle 
FT'D, is equal to FD'V. But FT'D is the exterior angle op¬ 
posite to FD'V; hence TT' is parallel to VV'. Therefore, 
tangents, &c. 

Cor. If tangents are drawn through the vertices of any 
two diameters, they will form a parallelogram. 



PROPOSITION VII. THEOREM. 

If through the vertex of any diameter , straight lines are 
drawn from the foci, meeting the conjugate diameter , the part 
intercepted by the conjugate is equal to half of the major axis. 

Let EE' be a diameter conjugate to 
DD', and let the lines DF., DF' be 
drawn, and produced, if necessary, so 
as to meet EE' in H and K ; then will 
DH or DK be equal to AC. 

Draw F'G parallel to EE' or TT', 
meeting FD produced in G. Then the 
angle DGF' is equal to the exterior 
angle FDT'; and the angle DF'G is 
equal to the alternate angle F'DT'. 

But the angles FDT', F'DT' are equal 
to each other (Prop. V.); hence the 
angles DGF', DF'G are equal to each other, and DG is equal 
to DF'. Also, because CK is parallel to F'G, and CF is equal 
to CF'; therefore FK must be equal to KG. 






HYPERBOLA. 


211 


Hence F'D — FD is equal toGD—FD or GF—2DF; that 
is, 2KF—2DF or 2DK. But F'D—FD is equal to 2AC. 
Therefore 2AC is equal to 2DK, or AC is equal to DK. 

Also, the angle DHK is equal to DKH ; and hence DH is 
equal to DK or AC. Therefore, if through the vertex, &c. 


PROPOSITION VIII. THEOREM. 

Perpendiculars drawn from the foci upon a tangent to the 
hyperbola , meet the tangent in the circumference of a circle 
whose diameter is the major axis. 

Let TT' be a tangent to the hyper¬ 
bola at D, and from F draw FE per¬ 
pendicular to TT' ; the point E will 
be in the circumference of a circle de¬ 
scribed upon AA' as a diameter. 

Join CE, FD, F'D, and produce FE 
to meet F'D in G. 

Then, in the two triangles DEF, 

DEG, because DE is common to both 
triangles, the angles at E are equal, be¬ 
ing right angles; also, the angle EDF is equal to EDG 
(Prop. Y.) ; therefore DF is equal to DG, and EF to EG. 

Also, because FE is equal to EG, and CF is equal to CF', 
CE must be parallel to F'G, and, consequently, equal to half 
of F'G. 

But, since DG has been proved equal to DF, F'G is equal 
to F'D — FD, which is equal to AA'. Hence CE is equal to 
half of AA' or AC ; and a circle described with C as a cen¬ 
ter, and radius CA, will pass through the point E. The same 
may be proved of a perpendicular let fall upon TT' from the 
focus F'. Therefore, perpendiculars, &c. 



PROPOSITION IX. THEOREM. 

The product of the perpendiculars from the foci upon a tan - 
gent , is equal to the square of half the minor axis. 

Let TT' be a tangent to the hyperbola at any point E, 
and let the perpendiculars FD, F'G be drawn from the foci; 
then will the product of FD by F'G, be equal to the square 
of BC. 

On AA' as a diameter, describe a circle ; it will pass 
through the points D and G (Prop. VIII.). Join CD, and 




212 


CONIC SECTIONS. 


produce it to meet GF' in D'. Then, 
because FD and F'G are perpendicu¬ 
lar to the same straight line TT', they 
are parallel to each other, and the al¬ 
ternate angles CFD, CF'D' are equal. 

Also, the vertical angles DCF, D'CF' 
are equal, and CF is equal to CF'. 

Therefore (Prop. VII., B. I.), DF is 
equal to D'F', and CD is equal to CD'; 
that is, the point D' is in the circum¬ 
ference of the circle ADA'G. 

Hence DFxGF' is equal to D'F'xGF', which is equal to 
A'F'xF'A (Prop. XXVIII., Cor. 2, B. IV.), which is equal 
to BC a (Prop. IV.). Therefore, the product, &c. 

Cor. The triangles FDE, F'GE are similar; hence 
FD : F'G : : FE : F'E; 

that is, perpendiculars let fall from the foci upon a tangent , are 
to each other as the distances of the point of contact from the 
Joci. 



PROPOSITION X. THEOREM. 

If a tangent and ordinate he drawn from the same point of 
an hyperbola , meeting either axis produced, half of that axis 
will be a mean proportional between the distances of the two in¬ 
tersections from the center. 

Let DTT' be a tangent to the 
hyperbola, and DG an ordinate to 
the major axis from the point of 
contact; then we shall have 
CT : CA : : CA : CG. 

Join DF, DF'; then, since the 
angle FDF' is bisected by DT 
(Prop. V.), we have 

F'T : FT : : F'D : FD 
(Prop. XVII., B. IV.). 

Hence, by Prop. VII., Cor., B. II., 

F'T—FT : F'T+FT : : F'D—FD : F'D+FD, 
or 2CT : F'F : : 2CA: F'D+FD ; 

that is, 2CT : 2CA : : F'F : F'D+FD. (1) 

Again, because DG is drawn from the vertex of the trian¬ 
gle FDF' perpendicular to the base FF' produced, we have 
(Prop. XXXI., Cor., B. IV.), 

F'G-FG : F'D+FD : : F'D—FD : F'G+FG, 

F'F : F'D+FD : : 2CA : 2CG. (2) 









HYPERBOLA. 


213 


Comparing proportions (1) and (2), we have 
2CT : 2CA : : 2CA : 2CG, 
or CT : CA : : CA : CG. 

It may also be proved that 

CT' : CB : : CB : CG'. 
Therefore, if a tangent, &c. 


PROPOSITION XI. THEOREM. 

The subtangent of an hyperbola , is equal to the correspond¬ 
ing subtangent of the circle described upon its major axis . 

Let AEA' be a circle described on 
A A' the major axis of an hyperbola; 
and from any point E in the circle, 
draw the ordinate ET. Through T 
draw the line DT touching the hyper¬ 
bola in D, and from the point of con¬ 
tact draw the ordinate DG. Join GE; 
then will GE be a tangent to the cir¬ 
cle at E. 

Join CE. Then, by the last Proposition, 

CT : CA : : CA : CG; 
or, because CA is equal to CE, 

CT : CE : : CE : CG. 

Hence the triangles CET, CGE having the angle at C 
common, and the sides about this angle proportional, are simi¬ 
lar. Therefore the angle CEG, being equal to the angle 
CTE, is a right angle ; that is, the line GE is perpendicular 
to the radius CE, and is, consequently, a tangent to the cir¬ 
cle (Prop. IX., B. III.). Hence GT is the subtangent cor¬ 
responding to each of the tangents DT and EG. Therefore, 
the subtangent, &c. 



PROPOSITION XII. THEOREM. 

The square of either axis , is to the square of the other , as the 
rectangle of the abscissas of the former, is to the square of their 
ordinate. 

Let DE be an ordinate to the major axis from the point 
D ; then we shall have 

CA 2 : CB 2 : : AExEA' : DE 2 . 

Draw DTT' a tangent to the hyperbola at D; then, by 
Prop. X., CT : CA : : CA : CE. 




214 


CONIC SECTIONS. 


Hence (Prop. XH., B. II.) 

CA 2 : CE 2 : : CT : CE; 
and, by division (Prop. VII., B. II.), 

CA 2 : CE 2 —CA 2 : : CT : ET. (1) 

Again, by Prop. X., 

CT' : CB : : CB : CE' or DE. 

Hence (Prop. XII., B. II.), 

CB 2 : DE 2 : : CT' : DE. 

But, by similar triangles, 

CT': DE : : CT : ET ; 

therefore CB 2 : DE 2 : : CT : ET. (2) 

Comparing proportions (1) and (2), we have 
CA 2 : CE 2 —CA 2 : : CB 2 : DE 2 . 

But CE 2 —CA 2 is equal to AE x EA' (Prop. X., B. IV.); hence 
CA 2 : CB 2 : : AExEA' : DE 2 . 

In the same manner it may proved that 

CB 2 : CA 2 : : BE'xE'B': D'E' 2 - 
Therefore, the square, &c. 

Cor . 1. CA 2 : CB 2 : : CE 2 -CA 2 : DE 2 . 

Cor. 2. The squares of the ordinates to either axis, are to 
each other as the rectangles of their abscissas. 

Cor. 3. If a circle be described on 
the major axis, then any tangent to 
the circle, is to the corresponding or¬ 
dinate in the hyperbola, as the major 
axis is to the minor axis. 

For, by the Proposition, 

CA 2 : CB 3 : : AExEA' : DE 2 . 

But AExEA' is equal to GE 2 (Prop. 

XXVIII., B. IV.). 

Therefore CA 2 : CB 3 : : GE 2 : DE 2 , 

or CA : CB : : GE : DE. 




PROPOSITION XIII. THEOREM. 


The latus rectum is a third proportional to the major and 
minor axes. 


Let LL' be a double ordinate to 
the major axis passing through the 
focus F ; then we shall have 

AA' : BB' : : BB' : LL'. 
Because LF is an ordinate to the ma¬ 
jor axis, 

AC 2 : BC 2 :: AF X FA': LF 2 (Prop. XII.) 
::BC 2 s LF 2 (Prop. IV.) 














HYPERBOLA. 

Hence 

AC : BC : : BC : LF. 

or 

AA' : BB' : : BB' : LI/. 

Therefore, the latus rectum, &c. 


PROPOSITION XIV. THEOREM. 


If from the vertices of two conjugate diameters , ordinates are 
drawn to either axis , the difference of their squares will he 
equal to the square of half the other axis. 

Let DD', EE' be any two conju¬ 
gate diameters, DG and EH ordinates 
to the major axis drawn from their 
vertices, in which case, CG and CH 
will be equal to the ordinates to the 
minor axis drawn from the same 
points ; then we shall have 
CA 2 =CG 2 —CH 2 , and CB 2 =EH 2 -DG 2 . 

Let DT be a tangent to the curve at 
D, and ET' a tangent at E. Then, by Prop. X., 

CGxCT is equal to CA 2 , or CHxCT'; 

whence 

CG : CH : : CT' : CT ; or, by similar triangles, 

: : CE : DT ; that is, 

: : CH : GT. 

Hence CH 2 =GTxCG 

= (CG—CT) x CG 
= CG 2 —CGxCT 
= CG 2 - CA 2 (Prop. X.) ; 
that is CA 2 =CG 2 —CH 2 . 

In the same manner it may be proved that 
CB 2 =EH 2 —DG 2 . 

Therefore, if from the vertices, &c. 

Cor. 1. CH 2 is equal to CG 2 — CA 2 ; that is, CG X GT; hence 
(Prop. XII., Cor. 1), 

CA 2 : CB 2 : : CGxGT : DG 2 . 

Cor. 2. CG 2 is equal to CA 2 +CH 2 ; hence 
CA 2 : CB 2 : : CG 2 : EH 2 . 








216 


CONIC SECTIONS. 


PROPOSITION XV. THEOREM. 

The difference of the squares of any two conjugate diameters , 
is equal to the difference of the squares of the axes. 

Let DD', EE' be any two conju¬ 
gate diameters ; then we shall have 
DD' 2 -EE' 2 =AA' 2 -BB'\ 

Draw DG, EH ordinates to the ma¬ 
jor axis. Then, by the preceding 
Proposition, 

CG 2 —CH 2 =CA 2 , 
and EH 2 —DG 2 =CB\ 

Hence CG 2 +DG 2 -CH 2 -EH 2 = CA 2 -CB 2 , 

or CD 2 —CE 2 =CA 2 -CB 2 ; 

that is, DD' 2 —EE' 2 =AA' 2 —BB'\ 

Therefore, the difference of the squares, &c. 



PROPOSITION XVI. THEOREM. 

The parallelogram formed by drawing tangents through the 
vertices of two conjugate diameters , is equal to the rectangle of 
the axes. 

Let DED'E' be a parallelogram, 
formed by drawing tangents to the 
conjugate hyperbolas through the 
vertices of two conjugate diame¬ 
ters DD', EE'; its area is equal to 
AA'xBB'. 

Let the tangent at D meet the 
major axis in T ; join ET, and draw 
the ordinates DG, EH. 

Then, by Prop. XIV., Cor. 2, we have 
CA 2 : CB 2 : : CG 2 : EH 2 , 
or CA : CB : : CG : EH. 

But CT : CA : : CA : CG (Prop. X.); 

hence CT : CB : : CA : EH, 

or CA x CB is equal to CTxEH, 

which is equal to twice the triangle CTE, or the parallelo¬ 
gram DE; since the triangle and parallelogram have the 
same base CE, and are between the same parallels. 

Hence 4CAxCB or AA'xBB' is equal to 4DE, or the 
parallelogram DED'E'. Therefore, the parallelogram, &c. 








HYPERBOLA. 


217 


PROPOSITION XVII. THEOREM. 

If from, the vertex of any diameter , straight lines are drawn 
to the foci , their 'product is equal to the square of half the con¬ 
jugate diameter. 

Let DD', EE' be two conjugate 
diameters, and from D let lines 
be drawn to the foci; then will 
FDxF'D be equal to EC 3 . 

Draw a tangent to the hyper¬ 
bola at D, and upon it let fall the 
perpendiculars FG, F'H; draw, 
also, DK perpendicular to EE'. 

Then, because the triangles 
DFG, DLK, DF'H are similar, 
we have 

FD : FG : : DL : DK. 

Also, F'D : F'H : : DL : DK. 

Whence (Prop. XI., B. II.), 

FD X F D : FG X F'H : : DL 2 ; DK 2 . (1) 
But, by Prop. XVI., AC x BC = EC x DK ; 
whence AC or DL : DK : : EC : BC, 

and DL 2 : DK 2 : : EC 3 : BC 2 . (2) 

Comparing proportions (1) and (2), we have 
FD x F'D : FG x F'H : : EC 2 : BC 2 . 

But FGxF'H is equal to BC 2 (Prop. IX.); hence FDxF'D 
is equal to EC 2 . Therefore, if from the vertex, &c. 



PROPOSITION XVIII. THEOREM. 

If a tangent and ordinate he drawn from the same point of 
an hyperbola to any diameter , half of that diameter will he a 
mean proportional between the distances of the two intersections 
from the center. 

Let a tangent EG and an ordinate EH be drawn from the 
same point E of an hyperbola, meeting the diameter CD 
produced; then we shall have 

CG : CD : : CD : CH. 

Produce GE and HE to meet the major axis in K and L; 
draw DT a tangent to the curve at the point D, and draw 
DM parallel to GK. Also, draw the ordinates EN, DO. 

Bv Prop. XIV., Cor. 1, CA 2 : CB 2 : : COxOT : DO 2 , 

: : CN x NK : EN 2 . 




218 


CONIC SECTIONS. 



Hence 

COxOT : CNxNK : : DO 3 : EN a 

: : OT 3 : NL 3 , by similar triangles. (1) 
Also, by similar triangles, OT : NL : : DO : EN 

: : OM : NK. (2) 
Multiplying together proportions (1) and (2) (Prop. XI., 
B. II.), and omitting the factor OT 3 in the antecedents, and 
NKxNL in the consequents, we have 

CO : CN : : OM : NL; 

and, by division, CO : CN : : CM : CL. (S) 

Also, by Prop. X., CKxCN=CA 3 =CTxCO; 
hence CO : CN : : CK : CT. (4) 

Comparing proportions (3) and (4), we have 
CK : CM : : CT : CL. 

But CK : CM : : CG : CD, 

and CT : CL : : CD : CH; 

hence CG : CD : : CD : CH. 

Therefore, if a tangent, &c. 


PROPOSITION XIX. THEOREM. 

The square of any diameter , is to the square of its conjugate , 
as the rectangle of its abscissas, is to the square of their ordinate. 

Let DD', EE' be two conju¬ 
gate diameters, and GH an or¬ 
dinate to DD'; then 
DD' 3 :EE' 3 ::DHxHD':GH 3 . 

Draw GTT' a tangent to the 
curve at the point G, and draw 
GK an ordinate to EE'. Then, 
by Prop. XVIII., 

CT : CD : : CD : CH, 
and CD 3 : CH 3 : : CT : CH 
(Prop. XII., B. II.), 





HYPERBOLA. 


219 


whence, by division, CD 2 : CH 2 -CD 2 : : CT : HT. (1) 
Also, by Prop. XVIII., CT' : CE : : CE : CK, 
and CE 2 : CK 2 : : CT' : CK or GH, 

: : CT : HT. (2) 

Comparing proportions (1) and (2), we have 

CD 2 : CE 2 : : CH 2 -CD 2 : CK 2 or GH 2 , 
or DD' 2 : EE' 2 : : DH x HD' : GH 2 . 

Therefore, the square, &c. 

Cor. 1. In the same manner it may be proved that DD /2 : 
EE' 2 : : DHxHD' : G'H 2 ; hence GH is equal to G'H, or 
every diameter bisects its double ordinates. 

Cor. 2. The squares of the ordinates to any diameter, are 
to each other as the rectangles of their abscissas. 


PROPOSITION XX. THEOREM. 

If a cone be cut by a plane , not passing through the vertex, 
and making an angle with the base greater than that made by 
the side of the cone , the section is an hyperbola. 

Let ABC be a cone cut by a plane 
DGH, not passing through the vertex, 
and making an angle with the base 
greater than that made by the side of 
the cone, the section DHG is an hyper¬ 
bola. 

Let ABC be a section through the axis 
of the cone, and perpendicular to the 
plane HDG. Let bgcd be a section 
made by a plane parallel to the base of 
the cone; then DE, the intersection of 
the planes HDG, BGCD, will be perpen¬ 
dicular to the plane ABC, and, consequently, to each of the 
lines BC, HE. So, also, de will be perpendicular to be and 
HE. Let AB and HE be produced to meet in L. 

Now, because the triangles LBE, L be are similar, as also 
the triangles HEC, Hec, we have the proportions 
BE i be i :*EL : eL 
EC : ec : : HE : He. 

Hence, by Prop. XI., B. II., 

BE X EC : be X ec : : HE X EL : He X eL. 

But, since BC is a diameter of the circle BGCD, and DE is 
perpendicular to BC, we have (Prop. XXII., Cor., B. IV.), 
BExEC = DE 2 . 

For the same reason, 

bexec—de \ 








220 


CONIC SECTIONS. 


Substituting these values of BE X EC and be, X ec, in the pre¬ 
ceding proportion, we have 

DE 2 : de 1 : : HE x EL : He x eh ; 
that is, the squares of the ordinates to the diameter HE, are 
to each other as the products of the corresponding abscissas. 
Therefore the curve is an hyperbola (Prop. XII., Cor. 2) 
whose major axis is LH. Hence the hyperbola is called a 
conic section , as mentioned on page 177. 


OF THE ASYMPTOTES. 

Definition. —An asymptote of an hyperbola is a straight 
line drawn through the center, which' approaches nearer the 
curve, the further it is produced, but being extended ever so 
far, can never meet the curve. 


PROPOSITION XXI. THEOREM. 

If tangents to four conjugate hyperbolas be drawn through 
the vertices of the axes , the diagonals of the rectangle so formed 
are asymptotes to the curves. 

Let AA', BB' be the axes of 
four conjugate hyperbolas, and 
through the vertices A, A', B, 

B', let tangents to the curve be 
drawn, and let CE, GE' be the 
diagonals of the rectangle thus 
formed; CE and CE' will be 
asymptotes to the curves. 

From any point D of one of the 
curves, draw the ordinate DG, 
and produce it to meet CE in H. 

Then, from similar triangles, we shall have 
CG 2 : GH 2 : : CA 2 : AE 2 or CB 2 , 

: : CG 2 —CA 2 : DG 2 (Prop. XII., Cor. 1). 

Now, according as the ordinate DG is drawn at a greater 
distance from .the vertex, CG 2 increases in comparison with 
CA ; that is, the ratio of CG 2 to CG 2 — CA 2 continually ap¬ 
proaches to a ratio of equality. But however much CG may 
be increased, CG 2 —CA 2 can never become equal to CG 2 ; 
hence DG can never become equal to HG, but approaches 
continually nearer to an equality with it, the further we re¬ 
cede from the vertex. Hence CH is an asymptote of the 
hyperbola; since it is a line drawn through the center, which 











HYPERBOLA. 


221 


approaches nearer the curve, the further it is produced, but 
being extended ever so far, can never meet the curve. 

In the same manner it may be proved that CH is an 
asymptote of the conjugate hyperbola. 

Cor. 1. The two asymptotes make equal angles with the 
major axis, and also with the minor axis. 

Cor. 2. The line AB joining the vertices of the two axes, is 
bisected by one asymptote, and is parallel to the other. 

Cor. 3. All lines perpendicular to either axis, and termi¬ 
nated by the asymptotes, are bisected by that axis. 


PROPOSITION XXII. THEOREM. 

If an ordinate to either axis he produced to meet the asymp¬ 
totes, the rectangle of the segments into which it is divided by 
the curve, will be equal to the square of half the other axis. 

Let DG be an ordinate to the 
major axis, and let it be produced 
to meet the asymptotes in H and 
H'; then will the rectangle HD X 
DH' be equal to BC 2 . 

For, by Prop. XII., Cor. 1, 

CA 2 : AE 2 : : CG a -CA 2 : DG 3 ; 
or, by similar triangles, 

: : CG 2 : GW 2 - 

Hence 

CG 2 : GH 2 : : CG 2 _CA 2 : DG 2 , 
and, by division, 

CG 2 : GH 2 : : CA 2 : GH 2 —DG 2 , or as CA 2 : AE 2 . 

Since the antecedents of this proportion are equal to each 
other, the consequents must be equal ; that is, 

AE 2 or BC 2 is equal to GH 2 —DG 2 ; 
which is equal to HDxDH'. 

So, also, it may be proved that 

CA 2 =D'K+D'L. 

Cor. HD x DH'=BC 2 = KM x MK'; that is, if ordinates to 
the major axis be produced to meet the asymptotes, the rect¬ 
angles of the segments into which these lines are divided by 
the curve, are equal to each other. 












222 


CONIC SECTIONS. 


PROPOSITION XXIII. THEOREM. 

All the 'parallelograms formed by drawing lines from any 
point of an hyperbola parallel to the asymptotes , are equal to 
each other . 

Let CH, CH' be the asymptotes 
of an hyperbola; let the lines AK, 

DL be drawn parallel to CH', and 
the lines AKA DL' parallel to CH ; 
then will the parallelogram CLDL' 
be equal to the parallelogram 
CKAK'. 

Through the points A and D 
draw EE', HH', perpendicular to 
the major axis; then, because the 
triangles AEK, DHL are similar, 
as also the triangles AE'K', DH'L', 
we have the proportions 

AK : AE : : DL : DH. 

Also, AK' : AE' : : DL' : DH'. 

Hence (Prop. XI., B. II.), 

AK x AK' : AE x AE' : : DLxDL' : DHxDH'. 

But, by Prop. XXII., the consequents of this proportion are 
equal to each other; hence * 

AK X AK' is equal to DL x DL'. 

But the parallelograms CA, CD being equiangular, are as 
the rectangles of the sides which contain the equal angles 
(Prop. XXIII., Cor. 2, B. IV.) ; hence the parallelogram CD 
is equal to the parallelogram CA. 

Cor. Because the area of the rectangle DLxDL' is con¬ 
stant, DL varies inversely as DL'; that is, as DL' increases, 
DL diminishes; hence the asymptote continually approaches 
the curve, but never meets it. The asymptote CH may, 
therefore, be considered as a tangent to the curve at a point 
infinitely distant from C. 



THE END. 








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